Answer: from the vertical, one should aim 86.6°
Explanation:
height of the center of object = 7.0 m - 0.05 m = 6.95 m
now let the bullet hits centre at point A height x meters from the ground
also let t be the time taken for the bullet to hit the object
so distance travelled by the target will be
d = h - x = 6.95 - x
now using the equation of motion
d = 1/2gt²
so 1/2gt² = 6.95 - x
x = 6.95 - 1/2gt² .........let this be equ 1
let angle of fire be ∅
so v(cos∅) × t = 100
our velocity v is 1200 ft/sec = 365.76 m/s
365.76(cos∅) × t = 100 ........equ 2
also vertical position of the bullet after t is
y = y₀ + c(sin∅)t - 1/2gt²
y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3
After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;
x = y
we substitute
equ 1 = equ 3
6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²
6.95 - 1 = 365.76(sin∅)t - 1/2gt² + 1/2gt²
5.95 = 365.76(sin∅)t
t = 5.96 / 365.76(sin∅)
now input the above equ into equ 2
365.76(cos∅) × 5.96/365.76(sin∅) = 100
5.95(cos∅)/sin∅ = 100
tan∅ = 5.95/100 = 0.0595
∅ = 3.40°
therefore from the vertical, one should aim (90° - 3.40°) = 86.6°
What does it mean if something is Periodic?
Answer:
In the context of chemistry and the periodic table, periodicity refers to trends or recurring variations in element properties with increasing atomic number. Periodicity is caused by regular and predictable variations in element atomic structure
Explanation:
The amount of force a muscle can produce is known as
25
60
20
1. How many servings can be made with 25 cups of pasta, 30 cups of marinara, and 5
cups of parmesan?
Answer:
60
Explanation:
i think it 60 because u plus 25 +30=55+5=60
The water from a fire hose is directed at a building . The water leaves the pipe at a speed of 25 m/s at an angle of 60 deg to the horizontal. If the building is 45 m away, at what height does it hit the building ?
Answer:
Approximately [tex]14\; \rm m[/tex] above the height of the fire hose, assuming that air resistance is negligible and that [tex]g = -9.81\; \rm m \cdot s^{-2}[/tex].
Explanation:
Consider the motion of water particles in two directions: vertical and horizontal.
Assuming that air resistance on the water particles is negligible.
Vertically, water particles would accelerate at [tex](-9.81\; \rm m \cdot s^{-2})[/tex] (towards the ground.)Horizontally, water particles would travel at a constant velocity.Initial velocity: [tex]v_0 = 25\; \rm m \cdot s^{-1}[/tex]. Angle of elevation: [tex]\epsilon = 60^\circ[/tex]. Calculate the initial velocity of these water particles in these two components:
Initial horizontal velocity: [tex]v_0 (\text{horizontal}) = v_0 \cdot \cos\left(\epsilon\right) \approx 12.5\; \rm m \cdot s^{-1}[/tex].Initial vertical velocity: [tex]v_0 (\text{vertical}) = v_0 \cdot \sin\left(\epsilon\right) \approx 21.7\; \rm m \cdot s^{-1}[/tex].How much time would it take for a water particle reaches the building from the hose? That particle needs to travel [tex]x(\text{horizontal}) = 45\; \rm m[/tex] at a constant horizontal speed of [tex]v(\text{horizontal}) \approx 12.5\; \rm m \cdot s^{-1}[/tex]. Therefore:
[tex]\displaystyle t = \frac{x(\text{horizontal})}{v(\text{horizontal})} \approx 3.6\; \rm s[/tex].
In other words, that water particle would be approximately [tex]3.6\; \rm s[/tex] into its flight when it hits the building. What would be its height? Assuming that the air resistance on that particle is negligible. The height of that water particle at time [tex]t[/tex] may be modeled using the SUVAT equation:
[tex]\displaystyle x(\text{vertical}) = \frac{1}{2}\, a \, t^{2} + \left[v_0(\text{vertical})\right]\, t[/tex],
where:
[tex]x(\text{vertical})[/tex] gives the height of the water particle (relative to where it was launched.)[tex]a = g = -9.81\; \rm m \cdot s^{-2}[/tex] (negative because gravitational acceleration points towards the ground.)[tex]v_0(\text{vertical}) \approx 21.7\; \rm m \cdot s^{-1}[/tex] (see above.)[tex]t \approx 3.6\; \rm s[/tex] (that's the time when the water particle hits the building.)Calculate the height of the water particle when it hits the building:
[tex]\begin{aligned}x(\text{vertical}) &= \frac{1}{2}\, a \, t^{2} + \left[v_0(\text{vertical})\right]\, t \\ &\approx \frac{1}{2} \times \left(-9.81\; \rm m \cdot s^{-2}\right) \times (3.6\; \rm s)^2 + 21.7\; \rm m \cdot s^{-1} \times 3.6\; \rm s \\ &\approx 14\; \rm m\end{aligned}[/tex].
What do seismic waves and sound waves have in common? [Seismic and sound waves]
A They're mechanical waves incorrect answer
B They're electromagnetic waves incorrect answer
C They're phonetic waves incorrect answer
D They're permanent waves
Answer:
D
Explanation:
They continously go back and fourth on the ray of movement making the tremendous exponent of the scientificalsource behind it
d is the answer
lkjhgfdeswwwwwwwwwww3e
A space vehicle is coasting at a constant velocity of 16.9 m/s in the y- direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.395 m/s^2 in the x direction. After 38.8 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off.
Find:
a. The magnitude.
b. The direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y- direction
Answer:
See attached file for the answer
a 100 foot cliff drops vertically to a lake if the angle of elevation from a swimmer to the top of the cliff is 54.6 how far is the swimmer from the clif
Answer:
Explanation:
Please see attached photo for explanation.
In the attached photo, X is the distance of the swimmer from the cliff.
Thus, we can obtain the value of x by using Tan ratio as illustrated below:
θ = 54.6°
Opposite = 100 foot
Adjacent = x =.?
Tan θ = Opposite /Adjacent
Tan 54.6 = 100/x
Cross multiply
x × Tan 54.6 = 100
Divide both side by Tan 54.6
x = 100 / Tan 54.6
x = 71.07 foot
Therefore, the swimmer is 71.07 foot from the cliff.
illustrated map can you guys answer this pls. i will
give you 20 points
A truck collides with a sports car in a high-speed head-on collision. . The mass of the truck and its contents is 5 times larger than the mass of the car and its contents. Select greater than, less than or equal to. During the collision, the magnitude of the change in momentum of the truck is _____ the magnitude of the change in momentum of the car. A: Greater than B: Less than C: Equal to
Answer:
A: Greater than
Explanation:
Momentum (P) implies product of the mass (m) and velocity (v) of a body.
P = mv
It is measured in kgm/s.
Thus the greater the mass of a body, the more its momentum even when moving at a slower velocity. The change in momentum of an object is called its impulse.
Let the mass of the car be represented by [tex]m_{c}[/tex] and that of the truck be represented by [tex]m_{T}[/tex].
Thus,
[tex]m_{T}[/tex] = 5[tex]m_{c}[/tex]
P of the truck = 5[tex]m_{c}[/tex]v
P of the car = [tex]m_{c}[/tex]v
Thus, during collision, the change in momentum of the truck is greater than that of the car.
Answer:
The correct option is B
Explanation:
This question seeks to test the knowledge of the change in momentum and deep understanding of the Newton's first law of motion which states that a substance will continue to be in a state of rest or constant motion unless acted upon by an external force.
The formula for change in momentum is mass multiplied by change in velocity. where change in velocity is final velocity minus the initial velocity.
The mass of the truck is already said to be 5 times larger, the change in velocity of the car will definitely be negative because the car will pushed back by the truck. For instance, if the initial velocity of the car is 50m/s and the final velocity backwards is 10m/s, the change in velocity will be (-10-50) which will equal (-60)
However, for the truck, the change in velocity will still remain positive for the final velocity because it is expected that the truck will still move forward despite the heads-on collision. Hence, if the initial velocity is 50m/s and the final velocity is 10m/s, the change in velocity will be (10-50) which will equal (-40).
Assuming the mass of the car is 1g. From the formula for change in momentum
The truck will be assumed to be (5 × 1) × -40 = -200
The car will be assumed to be 1 × -60 = -60
Hence, the magnitude of change in momentum of the truck will be lesser than the magnitude of the change in momentum of the car.
Two planes are about to drop an empty fuel tank. At the moment of release each plane has the same speed of 195 m/s, and each tank is at the same height of around 1.10 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15° above the horizontal (Plane A) and the other is flying at an angle of 15° below the horizontal (Plane B).
1. Find the:
a. magnitude.
b. direction of the velocity with which the fuel tank hits the ground if it is from plane A.
2. Find the:
a. magnitude
b. direction of the velocity with which the fuel tank hits the ground if it is from plane B.
In each part, give the direction as a positive angle with respect to the horizontal.
Answer:
376.71
Explanation:
1.a
To find the angle between them we add 15 +15=30°
Since they aren't perpendicular we use cosine law
[tex] \alpha = 180 - 30[/tex]
[tex] \sqrt{ {a}^{2} + {b}^{2} - 2ab \cos( \alpha ) } [/tex]
[tex] \sqrt{ {195}^{2} + {195}^{2} - 2 \times 195 \times 195 \times \cos(150) } [/tex]
=376.711
Tyler throws a baseball, which accidentally breaks a window in his neighbor's house. Which of the following represents the action force?
1.The window breaking.
2.The baseball accelerating.
3.Tyler's arm throwing the ball.
Answer: Tyler’s arm throwing the ball
Explanation:
Answer:
tyler's arm throwing the ball
Explanation:
i did this assignment
A soccer player heads the ball and sends it flying vertically upwards at a speed of 18.0 m/s . How high above the players ' head does the ball travel ?
Answer:
16.53 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 18.0 m/s.
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
The maximum height reached by the ball can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 18² – (2 × 9.8 × h)
0 = 324 – 19.6h
Rearrange
19.6h = 324
Divide both side by 19.6
h = 324 / 19.6
h = 16.53 m
Therefore, the maximum height reached by the ball is 16.53 m
A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s.
What is the final velocity after 8.0
A rocket is launched from a height of 3 m with an initial velocity of 15 m/s What is the maximum height of the rocket? When will this occur?
If no extra acceleration is added to the rocket, then its velocity at time t is
v = 15 m/s - g t
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.
Also, recall that
v² - u² = 2 a ∆x
where u is initial speed, v is final speed, a is acceleration, and ∆x is net displacement.
At the rocket's maximum height ∆x, the velocity is 0. So, the maximum height is
0² - (15 m/s)² = 2 (-g) ∆x
∆x = (15 m/s)² / (2 * (9.80 m/s²)) ≈ 11.48 m
But this assumes the rocket is launched from the ground. We're given that the rocket is launced from 3 m above the ground, so we need to add this to the height above. So the maximum height is closer to 14.48 m.
As mentioned before, this happens when vertical velocity is 0:
0 = 15 m/s - g t
t = (15 m/s) / (9.80 m/s²) ≈ 1.53 s
Select the only incorrect statement from the following:
a. When resistors are connected in parallel, the voltage across each resistor is the same
b. When resistors are connected in series, the potential difference across each resistor is the same
c. When resistors are connected in parallel, the total current from the voltage source is the sum of the currents in the individual resistors
d. The current flowing in all resistors in series across a battery is the same as the current from the battery
e. The voltage across all the resistors in series is the sum of the individual voltages across each resistor
Answer: b. When resistors are connected in series, the potential difference across each resistor is the same.
Explanation: When resistors are in series, current is the same in all of rhe them. According to the First Law of Ohm,
voltage (potential difference) = resistance * current.
Resistance is the proportionality constant between voltage and current.
If current is the same, to "keep" resistance constant, voltage has to vary.
So, when resistors are in series, current is the same and potential difference or voltage varies.
a) Show that the surface temperature of a star can be inferred from measurements of blackbodyfluxes at two different frequencies, even if the stellar radius and distance are unknown.
b) Explain why in practice this method does not work well if both frequencies are on theRayleigh-Jeans side of the spectrum,hνkT.
c) Derive a simple, approximate expression for the temperature when both measurements areon the Wien tail,hνkT.
d) Derive an expression for the star’s radius if a distance measurement is also available (e.g.,from parallax).
Answer:
The answers to the questions have been solved in the attachment.
Explanation:
The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.
A net force of 40 N south acts on an object with a mass of 20 KG. What is the objects acceleration
Answer:
The answer is 2 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the mass
We have
[tex]a = \frac{40}{20} = \frac{4}{2} \\ [/tex]
We have the final answer as
2 m/s²Hope this helps you
A motorcycle is driven horizontally straight off a 22-m tall cliff with a speed of 65
m/s. How far from the base of the cliff will the motorcycle land?
Answer:
137.8 m
Explanation:
The following data were obtained from the question:
Height (h) = 22 m
Initial velocity (u) = 65 m/s
Horizontal distance (s) =.?
Next, we shall determine the time taken for the motorcycle to get to the ground. This can be obtained as follow:
Height (h) = 22 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
22 = ½ × 9.8 × t²
22 = 4.9 × t²
Divide both side by 4.9
t² = 22/4.9
Take the square root of both side.
t = √(22/4.9)
t = 2.12 s
Finally, we shall determine the horizontal distance travelled by the motorcycle. This is illustrated below:
Initial velocity (u) = 65 m/s
Time (t) = 2.12 s
Horizontal distance (s) =.?
s = ut
s = 65 × 2.12
s = 137.8 m
Thus, the motorcycle will land at 137.8 m away from the cliff
Sugar is made of carbon, hydrogen, and oxygen atoms. Sugar is
A a compound because its parts cannot be separated through physical mea
B
an element because it is a pure substance whose parts cannot be separat
С
a mixture because it is made of different kinds of atoms.
D
a solution because it dissolves when mixed with water.
A 4500 kg car accelerates from rest to 45.0
km/h over a distance of 90 m. Find the net force
acting on the car.
please help
The car undergoes an acceleration a such that
(45.0 km/h)² - 0² = 2 a (90 m)
90 m = 0.09 km, so
(45.0 km/h)² - 0² = 2 a (0.09 km)
Solve for a :
a = (45.0 km/h)² / (2 (0.09 km)) = 11,250 km/h²
Ignoring friction, the net force acting on the car points in the direction of its movement (it's also pulled down by gravity, but the ground pushes back up). Newton's second law then says that the net force F is equal to the mass m times the acceleration a, so that
F = (4500 kg) (11,250 km/h²)
Recall that Newtons (N) are measured as
1 N = 1 kg • m/s²
so we should convert everything accordingly:
11,250 km/h² = (11,250 km/h²) (1000 m/km) (1/3600 h/s)² ≈ 0.868 m/s²
Then the force is
F = (4500 kg) (0.868 m/s²) = 3906.25 N ≈ 3900 N
An object of mass 10kg is accelerated upward at 2 m/s2. What force is required
Answer:
The answer is 20 NExplanation:
The force acting on an object given the mass and acceleration we use the formula
force = mass × accelerationFrom the question
mass = 10kg
acceleration = 2 m/s²
We have
force = 10 × 2
We have the final answer as
20 NHope this helps you
Find the maximum frequency for a non distorted sine-wave output voltage of 12-V peak with an Op-Amp that has a slew rate of 1.5 V/micro seconds
Answer:
The maximum frequency is approximately 20,000 Hz
Explanation:
Given;
peak output voltage, V = 12 V
slew rate, = 1.5 V/μs
The slew rate is given by;
slew rate = 2πfV
where;
f is the maximum frequency given by;
[tex]f = \frac{slew \ rate}{2\pi V}\\\\ f = \frac{1.5 }{\mu*2\pi *12}\\\\ f = \frac{1.5 }{10^{-6}*2\pi *12}\\\\f = 19891.79\\\\f = 20,000 \ Hz\\\\f = 20 \ kHz[/tex]
Therefore, the maximum frequency is approximately 20,000 Hz
A 6.16-g bullet is moving horizontally with a velocity of 342 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1155 g, and its velocity is 0.728 m/s after the bullet passes through it. The mass of the second block is 1517 g.
(a) What is the velocity of the second block after the bullet imbeds itself?
(b) Find the ratio of the total kinetic energy after the collisions to that before the collisions.
Answer:
a
The value is [tex]v = 0.8351 \ m/s [/tex]
b
The ratio is [tex]\frac{K_a}{K_b} = 5.94 11 *10^{-6}[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m_b = 6.16 \ g = 0.00616 \ kg[/tex]
The velocity is [tex]u_b = 342 \ m/s[/tex]
mass of the first block is [tex]m__{{B_1}}} = 1155 \ g = 1.155 \ kg[/tex]
The velocity of the first block after bullet passes is [tex]v__{{B_1}}} = 0.728 m/s[/tex]
The mass of the second block is [tex]m__{{ B_2}}} = 1517 g = 1.517 \ kg[/tex]
Gnerally according to the law of momentum conservation
[tex]m_b * u_b + m__{{B_1}}} * u__{{B_1}}} = m__{{B_1}}} * v__{{B_1}}} + v_b * m_b[/tex]
Here [tex]v_b[/tex] is the velocity of the bullet emerging from the first block
and [tex]u__{{B_1}}}[/tex] is zero because initial the first block was at rest
So
[tex]0.00616 * 342 + 1.155* 0 = v_b * 0.00616 + 1.155 * 0.728 [/tex]
[tex]v_b = 206.5 \ m/s [/tex]
Considering the second block
Gnerally according to the law of momentum conservation
[tex]m_b * v_b + m__{{B_2}}} * u__{{B_2}}} = [m__{{B_2}}} +m_b] v[/tex]
Here [tex]u__{{B_2}}}[/tex] is zero because initial the second block was at rest
=> [tex]0.00616 * 206.5 + 1.517* 0= [1.517 +m_b] v [/tex]
=> [tex]0.00616 * 206.5 = [1.517 +0.00616] v [/tex]
=> [tex]v = 0.8351 \ m/s [/tex]
The kinetic energy of the bullet before collision is
[tex]K_b = \frac{1}{2} * m * u_b^2[/tex]
=> [tex]K_b = 0.5 * 0.00616 * 342^2[/tex]
=> [tex]K_b = 360.2 \ J [/tex]
The kinetic energy of the bullet after collision is
[tex]K_a = \frac{1}{2} * m * v^2[/tex]
=> [tex]K_a = 0.5 * 0.00616 * 0.8351^2[/tex]
=> [tex]K_a = 0.00214 \ J [/tex]
Generally the ratio of the kinetic energy is mathematically represented as
[tex]\frac{K_a}{K_b} = \frac{0.00214}{360.2 }[/tex]
=> [tex]\frac{K_a}{K_b} = \frac{0.00214}{360.2 }[/tex]
=> [tex]\frac{K_a}{K_b} = 5.94 11 *10^{-6}[/tex]
2. How far can a dog travel if he can travel at 2.5 m/s for 5 seconds?
Answer:
[tex]\boxed {\tt 12.5 \ meters}[/tex]
Explanation:
We want to find how far the dog can travel, or the distance.
Distance can be found by multiplying the speed and the time.
[tex]d=s*t[/tex]
The dog travels 2.5 meters per second and travels for 5 seconds. Therefore,
[tex]s=2.5 \ m/s\\t= 5 \ s[/tex]
Substitute the values into the formula.
[tex]d=2.5 \ m/s \ * 5 \ s[/tex]
Multiply. Note that the seconds, or "s" will cancel each other out when multiplying.
[tex]d= 2.5 \ m * 5[/tex]
[tex]d=12.5 \ m[/tex]
The dog can travel 12.5 meters at 2.5 m/s for 5 seconds.
what would happen to the gravitational force between earth and the moon if the distance between them increased?
Answer:
The gravitation force would weaken.
Explanation:
It would weaken as they would have less of a effect on eachother. Additionally, the sun would most likely grab the moon and catapult it.. but we dont talk about it.
A test driver at Incredible Motors, Inc., is testing a new model car having a speedometer calibrated to read rather than . The following series of speedometer readings was obtained during a test run:
Time () 0 2.0 4.0 6.0 8.0 10 12 14 16
Velocity () 0 0 2.0 5.0 10 15 20 22 22
(a) Compute the average acceleration duringeach 2-s interval. Is the acceleration constant?A. Yes
B. No (b) Is it constant during any part of the test run?A. Yes
B. No
Answer:
(a) No, the acceleration is not constant
(b) Yes, it is constant between 6 s and 12 s
Explanation:
Given;
time of motion; t(s) = 0 2 4 6 8 10 12 14 16
velocity (m/s); v = 0 0 2 5 10 15 20 22 22
Average acceleration for t= (2,0) and v = (0,0)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{0-0}{2-0} = 0 \ m/s^2[/tex]
Average acceleration for t = (4, 2) and v = (2,0)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{2-0}{4-2} = 1.0 \ m/s^2[/tex]
Average acceleration for t = (6, 4) and v = (5,2)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{5-2}{6-4} = 1.5 \ m/s^2[/tex]
Average acceleration for t = (8, 6) and v = (10,5)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{10-5}{8-6} = 2.5 \ m/s^2[/tex]
Average acceleration for t = (10, 8) and v = (15,10)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{15-10}{10-8} = 2.5 \ m/s^2[/tex]
Average acceleration for t = (12,10) and v = (20,15)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{20-15}{12-10} = 2.5 \ m/s^2[/tex]
Average acceleration for t = (14,12) and v = (22,20)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{22-20}{14-12} = 1.0 \ m/s^2[/tex]
Average acceleration for t = (22, 22) and v = (16,14)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{22-22}{16-14} = 0 \ m/s^2[/tex]
(a) B. No (The acceleration is not constant)
(b) A. Yes (it is constant between 6 s and 12 s)
A straight line is drawn on the surface of a 18-cm-radius turntable from the center to the perimeter. A bug crawls along this line from the center outward as the turntable spins counterclockwise at a constant 45 rpm. Its walking speed relative to the turntable is a steady 3.4 cm/s. Let its initial heading be in the positive x-direction, where the x- and y-directions are fixed relative to the laboratory. As the bug reaches the edge of the turntable (still traveling at 3.4 cm/s radially, relative to the turntable) what are the x and y components of the velocity of the bug?
Answer:
v_r = 3.4 cm / s , v_t = 84.82 cm / s
Explanation:
The speed of the insect has two components, one radial and the other tangential,
The radial component is the speed with which the insect moves with respect to the plate
v_r = 3.4 cm / s
The tangential component is given by
v = w r
Let's reduce the angular velocity to the SI system
w = 45 rpm (2π rad / 1 rev) (1 min / 60 s) = 4.712 rav / s
let's calculate
v = 4.712 18
v_t = 84.82 cm / s
An unbalanced force of 15 N is applied to a 13 kg mass. What is the acceleration of the mass?
Answer:
the acceleration of an object equals the net force acting on it divided by its mass, or a = F m .
Explanation:
Which are consequences of smoking while pregnant
Answer:
Explanation:
Tobacco. Smoking during pregnancy increases the risk of health problems for developing babies, including preterm birth, low birth weight, and birth defects of the mouth and lip. Smoking during and after pregnancy also increases the risk of sudden infant death syndrome (SIDS).
Answer:
B I - T H deficits, S l D S, S T I L L . B I - T H
Explanation:
2021 edg .
An object is dropped from rest.
What is the acceleration after 9 s? The
acceleration of gravity is 9.8 m/s
2
.
Taha xain malai ..........hhdd