Answers
Answer:
Option (C) is correct.
Explanation:
Assuming that the hiker starts walking from the origin O as shown in the figure.
First, he walks 200 m west to point A (say), then he walks 100 m north to the final point B (say) as shown in the figure.
The final point B is in the north-west direction, therefore, the resulting point is in the north-west direction.
Hence, option (C) is correct.
Related Questions
Remaining Time: 1 hour, 49 minutes, 34 seconds.
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QUESTION 8
If the magnification is calculated to be +0.5 in a particular situation, what does that tell you about the image?
O A. It is bigger by 0.5
OB
It is half the size of the object.
O C. It is twice the size of the object.
O D. It is smaller by 0.5
O E. It tells you nothing about the image.
QUESTION 9
Answers
Explanation:
i think C . it is twice the size of the object
Assume the mass of a shuttlecraft is 2,200 kilograms [kg]. If the shuttlecraft requires 3 minutes [min] to rise from the surface of the Moon to an altitude of 19 kilometers [km], and the shuttle's engine is rated at 500 kilowatts [kW], what is the efficiency of the engines
Answers
Answer:
The efficiency of the engine is 75.94 %
Explanation:
Given;
mass of the shuttlecraft, m = 2,200 Kg
time required by the shuttlecraft, t = 3 minutes = 180 s
height risen by the shuttle, h = 19 km = 19,000 m
Input power of the shuttle, Pi = 500 kW = 500,000 W
The potential energy due to height risen by the Shuttle on Moon surface is given as;
E = mgh
where;
g is the acceleration due to gravity on moon, = ¹/₆ x 9.81 m/s² = 1.635 m/s²
E = 2200 x 1.635 x 19,000
E = 68343000 J
Output power of the shuttle, is given as Energy / time
Output power = (68343000) / (180)
Output power = 379,683.33 W
The efficiency of the shuttle is given as;
Efficiency = (Output power) / (Input power)
Efficiency = (379,683.33) / (500,000)
Efficiency = 0.7594
Efficiency(%) = 75.94 %
Therefore, the efficiency of the engine is 75.94 %
Jupiter is the largest planet in our solar system, having a mass and radius that are, respectively, 318 and 11.2 times that of earth. Suppose that an object falls from rest near the surface of each planet, and that each object falls the same distance before striking the ground. Determine the ratio of the time of fall on Jupiter to that on earth.
Answers
Answer:
0.62
Explanation:
we know that [tex]g=\frac{G M}{R^{2}}[/tex]
[tex]\frac{g_{g}}{g_{J}}=\frac{M_{g}}{M_{J}} \times \frac{R_{j}^{2}}{R_{z}^{2}}=\frac{M}{318 M} \times \frac{(11.2)^{2} R_{g}^{2}}{R_{g}^{2}}[/tex]
[tex]\frac{g_{g}}{g_{j}}=\frac{125.44}{318}=0.394[/tex]
We know that [tex]=\sqrt{\frac{2 h}{g}}[/tex]
here given that each object falls the same distance
[tex]\therefore t \alpha \sqrt{\frac{1}{g}}[/tex]
[tex]\therefore \frac{t_{J}}{t_{B}}=\sqrt{\frac{g_{g}}{g_{J}}}=\sqrt{\frac{g}{0.394}}[/tex]
[tex]\therefore \frac{t_{j}}{t_{g}}=\sqrt{0.394}=0.62[/tex]
A hockey player strikes a puck that is initially at rest. The force exerted by the stick on the puck is 975 N, and the stick is in contact with the puck for 0.0049 s.
(a) What is the impulse imparted by the stick to the puck.
___________ kg m/s
(b) What is the speed of the puck (m= 1.67 kg)just after it leaves the hockey stick?
____________ m/s
Answers
Explanation:
Given that,
The force exerted by the stick on the puck is 975 N
The stick is in contact with the puck for 0.0049 s
Initial speed of the puck, u = 0 (at rest)
(a) We need to find the impulse imparted by the stick to the puck.
Impulse = Force × time
J = 4.7775 kg-m/s
(b) Mass of the puck, m = 1.76 kg
We need to find the speed of the puck just after it leaves the hockey stick.
Let the speed be v.
As impulse is equal to the change in momentum.
[tex]J=m(v-u)\\\\4.7775=1.67(v-0)\\\\v=\dfrac{4.7775}{1.67}\\\\v=2.86\ m/s[/tex]
So, when the puck leaves the hockey stick its speed is 2.86 m/s.
7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent
beam 12cm from P. At what point does the beam converges if the lens is a) a convex lens of
focal length 20cm and b) a concave lens of focal length 16cm.
Answers
Answer:
Lens at a distance = 7.5 cm
Lens at a distance = 6.86 cm (Approx)
Explanation:
Given:
Object distance u = 12 cm
a) Focal length = 20 cm
b) Focal length = 16 cm
Computation:
a. 1/v = 1/u + 1/f
1/v = 1/20 + 1/12
v = 7.5 cm
Lens at a distance = 7.5 cm
b. 1/v = 1/u + 1/f
1/v = 1/16 + 1/12
v = 6.86 cm (Approx)
Lens at a distance = 6.86 cm (Approx)
earth has a radius of 6370 km what is earth's mass?
Answers
Answer:
approx 6 × 10^24 kg is earth's mass
. Which of the following is the best description about cancer cells?
Answers
Cancer cells are cells that divide relentlessly, forming solid tumors or flooding the blood with abnormal cells. Cell division is a normal process used by the body for growth and repair.
Is the amount of stretch of the springs proportional to the hanging mass? Explain briefly.
Derive a theoretical expression to find k-equivalent for springs in parallel and springs in series.
Answers
Answer:
When connected in parallel, the extension of both springs is the same, and the total elastic force will be equal to the sum of the forces in each spring: x=x1=x2,F=F1+F2. F=F1+F2,⇒kx=k1x1+k2x2=(k1+k2)x,⇒k=k1+k2.
Yes. The amount of stretch of a spring is proportional to the hanging mass.
According to Hooke's Law, the restoring force of the spring F is:
F = -kx
k is the spring constant and x is the stretch of spring.
The restoring force F is in opposite direction of the weight of the hanging mass which is mg
⇒ F = -mg
mg = -kx
x = mg/k
Hence x, the stretch of the spring is directly proportional to the hanging mass.
(i) If two springs are connected in parallel, their k- equialent is
[tex]k= k_{1}+k_{2}[/tex]
(ii) If two springs are connected in series, their k- equialent is
[tex]\frac{1}{k}=\frac{1}{k_{1} } +\frac{1}{k_{2} }[/tex]
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Consider two cars that are on course for a head-on collision. If they have masses m1 = 1,300 kg and m2 = 1,800 kg and are both traveling at 34 m/s, what is the magnitude of the total momentum?
Answers
Answer:
105400kgm/s
Explanation:
Momentum is the product of the mass and velocity of an object.
Momentum = mass × velocity
Given
Masses m1 = 1,300 kg and m2 = 1,800
velocity = 34m/s
Magnitude of the total momentum = (m1+m2)v
Magnitude of the total momentum = (1300+1800)((34)
Magnitude of the total momentum = 3100(34)
Magnitude of the total momentum = 105400kgm/s
An ideal gas is confined to a container with constant volume. The number of moles is constant. By what factor will the pressure change if the absolute temperature triples
Answers
Answer:
1/3
Explanation:
Gay Lusaac's law states that "the pressure of a given mass of gas is directly proportional with the absolute temperature of the gas, provided that the volume is kept constant."
In formula, we say that
P/T = k
Where
P = pressure at different points
T = temperature at different points
k = constant of proportionality
From the stated formula, if we multiply the temperature by 3, we have
P/3T = k
P * 1/3T = k
And from this, we see the pressure will change by a value of 1/3
In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?
A. BC shows zero acceleration, and AB shows negative acceleration.
B. AB shows zero acceleration, and CD shows negative acceleration.
C. BC shows zero acceleration, and CD shows negative acceleration.
D. AB shows zero acceleration, and BC shows negative acceleration.
Answers
Answer:
Option C. BC shows zero acceleration, and CD shows negative acceleration.
Explanation:
To successfully answer the question given above, we must know the meaning of zero acceleration and negative acceleration respectively.
Zero acceleration simply means the object is not accelerating. This implies that velocity of the object is constant (i.e unchanged) with time.
Negative acceleration simply means the object is decelerating i.e coming to rest.
Now, let us answer the question given above.
From the question given above, the following data were obtained:
1. AB indicates that the particle is accelerating because the velocity increased from A to B with time.
2. BC indicates zero acceleration because the velocity of the Particle is constant (i.e unchanged) with time.
3. CD indicates that the object is coming to rest i.e decelerating (negative acceleration) because the velocity of the object decreased from C to D with time.
From the above illustrations, we can see that only option C gives the correct answer to the question.
Answer:
Option C. BC shows zero acceleration, and CD shows negative acceleration.
Explanation:
My dad gifted me a calculator. I have observed that very small cells are used in a calculator. What are these cells called and what are their main advantage?
Answers
*WILL MARK BRAINLIEST*
A glass window (n = 1.52) has a
uniform layer of ice on it (n = 1.31).
What is the critical angle for a ray
trying to pass from glass to ice?
(Water n = 1.33, Air n = 1.00)
(Unit = deg)
Answers
Answer:
1.52
Explanation:
Answer: 59.5 deg
Explanation:
I got it right on acellus ✅
Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects that orbit the Sun –Not large enough to have enough gravitational pull to clear the path of their orbit –4 known in the Kuiper Belt (Eris, Haumea, Makemake, and Pluto) –1 in the asteroid belt (Ceres) Jasmine later copied the title of the notes from a friend. What title did Jasmine most likely copy? Comets The Kuiper Belt Asteroids Dwarf Planets
Answers
Answer:
D. Dwarf Planets
Explanation:
Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects that orbit the Sun –Not large enough to have enough gravitational pull to clear the path of their orbit, Jasmine is most likely to copy the Dwarf Planets, therefore the correct answer is option D.
What is a solar system?It is a system that collection of all the planets and spatial bodies revolving around the sun because of the gravitational pull of the sun.
Our Solar System is based on a heliocentric model in which the Sun is assumed to reside at the central point of the planetary system.
As given in the problem Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects that orbit the Sun –Not large enough to have enough gravitational pull to clear the path of their orbit –4 known in the Kuiper Belt (Eris, Haumea, Makemake, and Pluto) –1 in the asteroid belt (Ceres) Jasmine later copied the title of the notes from a friend.
Thus, Jasmine is most likely to copy the Dwarf Planets, therefore the correct answer is option D.
Learn more about Solar Systems here, refer to the link given below ;
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I’m in a test and I’m sort of being timed here’s a photo of question
Answers
Answer: D.
Explanation:
Answer:
i think the answer is b
Explanation:
playing sports often require both
A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted by the seat on the driver if the mass of the driver is 60 kg?
Answers
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
[tex]F_c = F_g - F_N[/tex]
where;
Fc is the centripetal force
[tex]F_g[/tex] is downward force due to weight of the driver
[tex]F_N[/tex] is upward or normal force on the drive
[tex]F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N[/tex]
Therefore, the normal force the seat exerted on the driver is 125 N.
The normal force exerted by the seat on the driver if the mass of the driver is 60 kg is of 125 N.
Given data:
The mass of car is, m' = 2000 kg.
The speed of car is, v = 100 km/h = 100 × 5/18 = 27.77 m/s.
The radius of curvature of path is, r = 100 m.
The mass of driver is, m = 60 kg.
In this case, the normal force on the driver is equal to the difference between weight of the driver and the centripetal force on the driver. Then the expression is given as,
[tex]N'= W - F\\\\N '= mg-\dfrac{mv^{2}}{r}[/tex]
Solving as,
[tex]N' = (60 \times 9.8)-\dfrac{60 \times 27.77^{2}}{100}\\\\N' = 125\;\rm N[/tex]
Thus, we can conclude that the normal force exerted by the seat on the driver if the mass of the driver is 60 kg is of 125 N.
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Is Chocolate Milk a compound,mixture or a element
Answers
Answer:
mixture
Explanation:
because of many Ingredients it is called a mixture
Solubility Curve Practice Problems Worksheet 1
Directions: Find the mass of solute will dissolve in 100mL of water at the following temperatures?
150
1. KCl at 70°C =
140
2. Ce,(SO.), at 100°C=
130
120
3. NH.Cl at 90°C=
4. Which of the above three substances is most
110
100
NaNO3
soluble in water at 15°C. =
90
KNO3
Grams of solute
per 100 g H,0
80
70
60
NH3
NHACI
50
KCI
40
Naci
30
20
KCIO
10
Ce2(SO4)3
0
10 20 30 40 50 60 70 80 90 100
Temperature (°C)
Answers
According to the solubility curve, KCl is soluble in water at a rate of about 40 grammes per 100 g. Given that we have 100 mL, or 100 grammes, of water, the mass of KCl that will dissolve is similarly 40 grammes.
[tex]Ce_2(SO_4)_3[/tex] is soluble in water at a rate of around 30 grammes per 100 grammes at 100 °C. [tex]Ce_2(SO_4)_3[/tex] will therefore dissolve in 100 mL of water at a mass of 30 grammes.
[tex]NH_4Cl[/tex] dissolves in water at a rate of around 90 grammes per 100 grammes at 90 °C. The mass of [tex]NH_4Cl[/tex] that will dissolve in 100 mL of water is similarly 90 grammes.
We must ascertain the solubilities of the abovementioned chemicals at 15°C in order to ascertain which substance is most soluble at that temperature.
[tex]NaNO_3[/tex] has the highest solubility at 15°C, as seen by the solubility curve. At 15°C, [tex]NaNO_3[/tex] dissolves in water at a rate of about 80 grammes per 100 g.
Therefore, at 15°C, NaNO3 is the most soluble substance among the three given substances.
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An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If there is a charge concentration of 40 C at height 3000 m with the cloud and -40 C at height 1000 m, what is the electric field at the aircraft
Answers
Answer:
[tex]400000\ \text{N/C}[/tex]
Explanation:
[tex]q_1[/tex] = Charge at 3000 m = 40 C
[tex]q_2[/tex] = Charge at 1000 m = -40 C
[tex]r_1[/tex] = 3000 m
[tex]r_2[/tex] = 1000 m
k = Coulomb constant = [tex]9\times10^9\ \text{Nm}^2/\text{C}^2[/tex]
Electric field due to the charge at 3000 m
[tex]E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}[/tex]
Electric field due to the charge at 1000 m
[tex]E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}[/tex]
Electric field at the aircraft is [tex]E_1+E_2=40000+360000=400000\ \text{N/C}[/tex].
What is the acceleration down the ramp of an object on a 17° ramp?
A
2.9 m/s2
B
4.2 m/s2
C
5.5 m/s2
d
9.4 m/s2
Answers
convert 144 km/h to m/s
Answers
Answer:
40m/s
Explanation:
144km/h
1km=1000m
1hr=3600secs
144×1000/3600=
40m/s
1. The center of gravity of both cats and humans is roughly in the thoracic region. However, they each bear their weight very differently. a. Which are the major, weight-bearing structures of the two skeletons
Answers
Answer:
Major, weight-bearing structures are the bones of the body that are strong and dense to be able to bear the weight of the body. The major, weight-bearing structures of cat and human skeletons are :
Human skeleton: The body weight of an individual is on his pelvic girdles that are attached to the bones of lower limbs. Thigh bones, leg bones, and bones of feet comprise lower limbs The lower limbs consist of the thigh, the leg, and the foot.
Cat skeleton: cats are quadrupedal so it bears all the body weight on shoulders and legs that includes the Scapula and pelvis.
A 15.0kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 degrees above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3.
a. Draw a free body diagram. Draw your coordinate system and label the axes.
b. Calculate the work done on the block by the 70 N force.
c. Calculate the work done on the block by the normal force.
d. Calculate the work done on the block by the gravitational force.
e. Calculate the work done on the block by the force of friction.
Answers
Answer:
[tex]W=70 * 5cos20 = 328.89 J[/tex]
[tex]W_n = 0[/tex]
[tex]W_g=0[/tex]
[tex]W_f= -184.59J[/tex]
Work done is 0
Explanation:
From the question we are told that
Weight of block =15.0kg
Force acting on the block = 70.0N
At an angle of 20 degree
Displacement of block is 5m
Coefficient of kinetic friction 0.3
b) Generally work done by force is give by [tex]W=fdcos \theta[/tex]
therefore
[tex]W=70 * 5cos20 = 328.89 J[/tex]
c) there is no work done by the normal force in this scenario because
normal force in this case is perpendicular to the displacement of the motion
[tex]W_n = 0[/tex]
d) The displacement in the vertical direction is 0
Therefore the gravitational work done is 0 [tex]W_g=0[/tex]
e)Generally in finding work done by friction we first find frictional force
Mathematically the equation for frictional force is given [tex]f = \alpha N[/tex]
Given that
[tex]N=mg-Fsin20[/tex]
[tex]N= 15.0*9.8 - 70 sin20[/tex]
[tex]N=123 N[/tex]
[tex]f=0.3* 123.06 = 36.92N[/tex]
Mathematically solving to get work done by frictional force [tex]W_f[/tex]
[tex]W_f= -fd\\W_f = -36.92 * 5[/tex]
[tex]W_f= -184.59J[/tex]
the frictional force work done is [tex]W_f= -184.59J[/tex]
Answer all these questions correctly and will mark brainliest!
1. In Football a field goal is kicked for how many points?
2. Football a touch down is worth how many points?
3. What part of the body can a goalie in soccer use that no one else can?
4. How many quarters are in a football game?
5. How many players are on the field in football?
6. Who invented basketball?
7. Who invented the game Nukem?
8. In volleyball can a ball (Including serve) be played off of the net?
9. When did basketball become an Olympic sport?
10.How many quarters are in a football game?
Inappropriate answers will be reported
Answers
Answer:
1. 3 points
2. 6 points
3. goalkeeper
4. four quarters
5. eleven players
6. James Naismith
7. Duke nukem ( I'm not sure)
8. no
9. 1936
10. four
HOPE IT'S HELP :)
What is the difference between static frication and kinetic friction
Answers
static friction is acting on stationery object (rest) but kinetic friction acting on a moving body.
The other one is static friction oppose the object to start a motion so it force is great than kinetic friction
Two particles, an electron and a proton, move in a circular path in a uniform magnetic field of intensity B=1.23 T. Find the ratio between the time period of the proton Tp to the electron Te (i.e., find 'p)
Answers
Answer:
The ratio of the time period of the proton to the electron is 1835.16.
Explanation:
Given that,
Two particles, an electron and a proton, move in a circular path in a uniform magnetic field of intensity B=1.23 T
We need to find the ratio between the time period of the proton Tp to the electron Te.
The time period in magnetic field is given by :
[tex]T=\dfrac{2\pi m}{qB}[/tex]
For proton, time period is :
[tex]T_P=\dfrac{2\pi m_P}{q_pB}\ ....(1)[/tex]
For an electron, the time period is :
[tex]T_e=\dfrac{2\pi m_e}{q_eB}\ ....(2)[/tex]
From equation (1) and (2) :
[tex]\dfrac{T_p}{T_e}=\dfrac{\dfrac{2\pi m_p}{q_pB}}{\dfrac{2\pi m_e}{q_eB}}\\\\As\ q_e=q_p\\\\\dfrac{T_p}{T_e}=\dfrac{m_p}{m_e}\\\\=\dfrac{1.67\times 10^{-27}}{9.1\times 10^{-31}}\\\\=1835.16[/tex]
So, the ratio of the time period of the proton to the electron is 1835.16.
alpha and beta rays origin is same
Answers
Answer:
Alpha particles carry a positive charge, beta particles carry a negative charge, and gamma rays are neutral. An alpha particle is made up of two protons and two neutrons bound together. Beta particles are high energy electrons. Gamma rays are waves of electromagnetic energy, or photons.
Determine whether each of the following statements is true or false.
a. An object's weight is always equal to its mass.
b. The force of tension always pushes.
c. The magnitude of the sum of the forces on an object is never greater than its weight. Explain.
Answers
Answer:
a) For an object with mass M, in a region with a gravitational acceleration g, is:
W = M*g
Then the weight is g times the mass of the object, this means that the weight is not always equal to the mass of the object, this first statement is false.
For example, in Earth the gravitational acceleration is 9.8m/s^2
Then there is no object in Earth with a weight equal to its mass.
b) The force of tension can be the force in a piece of string that is holding an object with mass M.
The force of tension will be always pointing in the direction to the "center" of the string, then it does not push, the tension force "pulls"
The statement is false.
c) we know that the weight is:
W = M*g
This is a force, that for an object that is in the air, will pull the object back to the ground.
Suppose that we also have that object attached to a string, and the object is in the air, now the object starts to fall due to its weight and we also pull down with the string, then the total force pulling down will be the weight plus the tension of the string, then we will have a force larger (in magnitude) than the weight, which means that the statement is false.
What is the strength of the electric field in a region where the electric potential is constant?
Answers
Answer:
Where the electric potential is constant, the strength of the electric field is zero.
Explanation:
As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e
Eₓ = - dV / dx ----------(i)
From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.
Therefore, a constant electric potential means that electric field is zero.
An ideal step-down transformer is needed to reduce a primary voltage of 120 V to 6.0 V. What must be the ratio of the number of turns in the secondary to the number of turns in the primary
Answers
Answer:
[tex]N_s : N_p = 20 : 1[/tex]
Explanation:
From the question we are told that
The primary voltage is [tex]V_p = 120 \ V[/tex]
The secondary voltage is [tex]V_s = 6 \ V[/tex]
Generally from the transformer equation we have that
[tex]\frac{V_p}{V_s} = \frac{N_p}{N_s}[/tex]
So
[tex]\frac{120}{6} = \frac{N_p}{N_s}[/tex]
=> [tex]\frac{N_p}{N_s} = 20[/tex]
Therefore the ratio of the number of turns in the secondary to the number of turns in the primary is
[tex]N_s : N_p = 20 : 1[/tex]