David drove the first 6 hours of his journey at 65km/hr and the last 3 hours of his journey at 80km/hour. How far is the whole journey in km?

Answer:

At 40 deg  Vy = V sin 40

at 70 deg Vy = V sin 70

So the ball launched at 70 deg has the greatest vertical velocity and will remain in the air the longest:

Since t = Vy / g    time for to reach zero vertical velocity and also the time for the ball to reach velocity Vy on the downward path

Mathis kicked a ball on a level surface at 30∘ and 60∘ with the same total speed as shown below.

Which launch angle results in the greater maximum height for the ball?

Answer: CORRECT (SELECTED)

60

Answer:

volume about 1.3 million Earths could fit inside the Sun the mass of the Sun is 1.989 X 10 to the 30 kg about 333000 times the mass of the Earth

Explanation:

step-by-step explanation hope this answer your question

Answer:

65 miles per hour.

Explanation:

Answer:

Average speed of entire trip is : 3.67 m/s

Explanation:

Speed is calculated as the distance covered divided by time

Average speed will be the :speed before stop add to speed after stopping then divide by 2

Speed before stop = 200/ 60 = 3.33 m/s

Speed after stop = 400/100= 4m/s

Average speed = (3.33+4)/2 = 7.33/2 = 3.67 m/s

Answer:

The distance will be x = 1320 [m]

Explanation:

To solve this problem we must use the expression of physics that relates space to time, which is defined as speed.

v = Speed = 11 [m/s]

t = time = 120 [s]

x = distance [m]

v = x/t

x = v*t

x = 11*120

x = 1320 [m]

Answer:

Pls ezplain your question more

Explanation:

Answer:

orbit is the answer

Answer:

Force=Mass*acceleration

on earth, acceleration=9.81 m/s^2

900 N=Mass*9.81 m/s^2

Mass=91.74 Kg

F=Mass*acceleration(Jupiter)

F=91.74Kg*25.9m/s

F=2376.066 N on Jupiter

Plz mark me as brainliest if u found it helpful

Complete Question

A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?

[tex]A.\ \ P_2=4P_1[/tex]

[tex]B.\ \  P_2=\frac{4}{3} P1[/tex]

[tex]C.\ \  P_2=P_1[/tex]

[tex]D. \ \ P_2=\frac{3}{4} P_1[/tex]

[tex]E. \ \ \ P_2=\frac{1}{3} P_1[/tex]

Answer:

The correct option is  B

Explanation:

From the question we are told that  

    The mass of the brick is  M

    The  height height of the 10th floor is  H =  4y

     The height attained during the first part of the lift is  [tex]h_1 =  3y[/tex]

     The time taken is [tex]t_1 =  4T[/tex]

    The height attained during the second part of the lift is  [tex]h_2  = y[/tex]

    The time taken is  [tex]t_2  =  T[/tex]

Generally the velocity of the crane during the first lift is mathematically represented as

         [tex]v_1  =  \frac{h_1}{t_1}[/tex]

=>      [tex]v_1  =  \frac{3y}{4T}[/tex]

Generally the velocity of the crane during the first lift is mathematically represented as

         [tex]v_1  =  \frac{h_2}{t_2}[/tex]

=>      [tex]v_1  =  \frac{y}{T}[/tex]

Generally the power generated during the first lift is  

     [tex]P_1 =  F_1 *  v_1[/tex]

Here [tex]F_1[/tex] force applied during the first lift which is mathematically represented as

       [tex]F_1  =  M  *  g[/tex] here g is the acceleration  due to gravity

So

       [tex]P_1 =  Mg * \frac{3y}{4T}[/tex]

Generally the power generated during the second lift is  

     [tex]P_2 =  F_2 *  v_2[/tex]

Here [tex]F_2[/tex] force applied during the second lift which is mathematically represented as

       [tex]F_2  =  M  *  g[/tex] here g is the acceleration  due to gravity

So

       [tex]P_2 =  Mg * \frac{y}{T}[/tex]

So the ratio  of the first power to the second power is  

      [tex]\frac{P_1}{P_2}  =  \frac{Mg * \frac{3y}{4T}[}{Mg * \frac{y}{T}}[/tex]

=>   [tex]\frac{P_1}{P_2}  = \frac{3}{4}[/tex]

=>    [tex]P_2 = \frac{4}{3} P_1[/tex]

Answer:

Explanation:

speed (velocity) = distance / time

                          = 0.2 km (1000 m/km)

                                     95 secs.

                          = 2.10 meters per second

Answer:

Single Replacement

Explanation:

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = [tex]q_{int}[/tex] /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

Answer:

It says the answer. (5 marks.)

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

Answer:

36m/s

Explanation:

v=u+at

v=21+(2.2×6.9)

v=21+15.18

v=36.18

v=36m/s

Answer:

[tex]Velocity=\frac{dx}{dt}[/tex]

Explanation:

Remember that instantaneous velocity is just a measure to know the velocity that any object has at any point given in time, so we just need to know the distance it has travel, which would be the change in position, and the time it took that change in position to occurr, this means distance by time, so we just divide dx by dt and we have the solution for instantaneous velocity.

If a change in position as denoted by [tex]dx[/tex] and [tex]dt[/tex] change in time, the instantaneous velocity will be given by,

[tex]v = \dfrac {dx}{dt}[/tex]

[tex]v = \dfrac {dx}{dt}[/tex]

Where,

[tex]v[/tex] - instantaneous velocity

[tex]dt[/tex]   - change in distance (position)

[tex]dt[/tex]- change in time

Therefore, if a change in position as denoted by [tex]dx[/tex] and [tex]dt[/tex]change in time, the instantaneous velocity will be given by,

[tex]v = \dfrac {dx}{dt}[/tex]

Learn more about  Instantaneous velocity.

https://brainly.com/question/13372043

Answer:

one right angel = 90 degrees

and two acute angels?

Explanation:

Answer:

Since the ball was thrown horizontally, there was no vertical component in that force. and hence, the initial vertical velocity of the ball is 0 m/s and the initial horizontal velocity is r.

We are given:

initial velocity  (u) = 0 m/s     [vertical]

final velocity (v) = v m/s  [vertical]

time taken to reach the ground (t) = t seconds

acceleration (a) = 10 m/s/s   [vertical , due to gravity]

height from the ground (h) = 130 m

displacement (s) = 53 m [horizontal]

Solving for time taken:

From the third equation of motion:

s = ut + 1/2 at²

130 = (0)(t) + 1/2 * (10) * t²

130 = 5t²

t² = 26

t = √26 seconds  or   5.1 seconds

Final Horizontal velocity of the ball

Since the horizontal velocity of the ball will remain constant:

the ball covered 53 m in 5.1 seconds [horizontally]

horizontal velocity of the ball = horizontal distance covered / time taken

Velocity of the ball = 53 / 5.1

Velocity of the ball = 10.4 m/s

Answer:

51.51519 m/s

Explanation:

Given: [tex]a_{x} =0[/tex] [tex]a_{y} -g[/tex] [tex]v_{yo} =0[/tex] [tex]x_{o} =0[/tex] [tex]x=53[/tex][tex]y_{o} =130[/tex]

X-direction                           | Y-direction

[tex]x=x_{o} +v_{xo}t[/tex]                         | [tex]y=y_{o} +v_{yo}t+\frac{1}{2}a_{y}t^2[/tex]

[tex]53=0v_{xo}(5.15078)[/tex]                 | [tex]0=130+\frac{1}{2}(-9.8)t^2[/tex]

[tex]53=v_{xo} (5.15078)[/tex]                    | [tex]-130=-4.9t^2[/tex]

[tex]\frac{53}{5.15078} =v_{xo}[/tex]                             |  [tex]\sqrt{\frac{-130}{-4.9} }=\sqrt{t^2}[/tex]

[tex]10.2897=v_{xo}[/tex]                            | [tex]5.15078=t[/tex]

[tex]v=\sqrt{v_{y}^2+ v_{x}^2}[/tex]                            | [tex]v_{y}^2 =v_{yo}+2a_{y} d[/tex]

[tex]v=\sqrt{(50.27771)^2+(10.2897)^2}[/tex] | [tex]\sqrt{v_{y}^2} =\sqrt{2(-9.8)(0-130)}[/tex]

[tex]v=51.51519 m/s[/tex]                        | [tex]v_{y}=50.47771[/tex]

the missing word is clockwise moment. I hope this helps good luck

Answer:

when gravitational force suddenly disappears, then only centrifugal force will be acting and velocity is tangential to the orbit and hence, the satellite will fly off tangentially with same velocity v.

Answer:

Usbe

Explanation:

Answer:

a) F = 9.69 N

b) F = 10.88 N

c) F = 8.51 N

Explanation:

a) The kinetic frictional force when the elevator is stationary is the following:

[tex] F_{k} = \mu_{k}N = \mu(mg) [/tex]    

Where:

F(k) is the kinetic frictional force

N is the normal force = mg

m: is the mass = 2.60 kg

g: is the gravity = 9.81 m/s²

μ(k) is the coefficient of kinetic friction = 0.380

[tex] F_{k} = \mu(mg) = 0.380*2.60 kg*9.81 m/s^{2} = 9.69 N [/tex]    

b) When the elevator is accelerating upward with acceleration "a" equal to 1.20 m/s²

[tex] F_{k} = \mu_{k}N = \mu[m(g + a)] [/tex]    

The normal is equal to mg plus ma because the elevator is accelerating upward

[tex] F_{k} = \mu m(g + a) = 0.380*2.60 kg(9.81 m/s^{2} + 1.20 m/s^{2}) = 10.88 N [/tex]    

c) When the elevator is accelerating downward with a =1.20 m/s² we can find the kinetic frictional force similar to the previous case:

[tex] F_{k} = \mu m(g - a) = 0.380*2.60 kg(9.81 m/s^{2} - 1.20 m/s^{2}) = 8.51 N [/tex]    

I hope it helps you!

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

  A

2

A

Explanation:

From the question the data given for Error (mV) is -15

-15.17

8.67

-13.74

-20.69

-6.96

-1.36

-2.96

-9.26

3.11

-14.12

6.39

-14.77

Generally

The null hypothesis is [tex]H_o : \mu = 0[/tex]

The alternative hypothesis is [tex]H_a : \mu \ne 0[/tex]

The sample size is n = 13

Here [tex]\mu[/tex] represents the true error bias (i.e population error bias)

Generally the sample error bias is mathematically represented as

[tex]\= x = \frac{ \sum x_i}{n}[/tex]

=> [tex]\= x = \frac{ -15.17 + 8.67 + (-13.74) + \cdots + (-14.77) }{13}[/tex]

[tex]\= x = -7.37[/tex]

Generally the standard deviation is mathematically represented as

[tex]\sigma = \sqrt{\frac{\sum (x_i - \= x )^2}{n} }[/tex]

=> [tex]\sigma = \sqrt{\frac{ (-15.17-( -7.37) )^2 + (8.67 -( -7.37) )^2 + \cdots + (-14.77 -( -7.37) )^2 }{13} }[/tex]

=> [tex]\sigma = \sqrt{ 119.385}[/tex]

=> [tex]\sigma = 10.926[/tex]

Generally the test statistics is mathematically represented as

[tex]t = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]

=> [tex]t = \frac{ -7.37 - 0 }{\frac{10.926}{\sqrt{13} } }[/tex]

=> [tex]t = -2.838[/tex]

Generally the p-value is mathematically represented as

[tex]p-value = 2 P(t < -2.432)[/tex]

From the z-table  [tex]P(t < -2.432) =  0.0075 [/tex]

So  [tex]p-value  =  2* 0.0075 [/tex]

=>  [tex]p-value  = 0.015 [/tex]

So given that  p-value is  less than the [tex] \alpha = 0.05[/tex] then we reject the null hypothesis and conclude that the oscilloscope has an error bias

Explanation:

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Answer:

25.88 N

Explanation:

Mass of the shopping cart,[tex]m=65 kg[/tex]

The coefficient of friction between the cart and the floor, [tex]\mu= 0.01[/tex]

Let, F be the required force to accelerate the cart at [tex]a=0.30 m/s^2[/tex].

Gravitational force, mg, acts downward which is being balanced by the normal reaction, N, on the cart by the floor.

As the motion of the cart in the vertical direction is zero. So, using the static equilibrium condition will be zero.

From free body diagram (FBD):

[tex]N-mg=0[/tex]

[tex]\Rightarrow N=mg.[/tex]

the net force action in this direction The frictional force, f, acts in the direction to opposes the motion of the cart as shown.

[tex]f=\muN=\mu mg[/tex]

Now, apply the dynamic equilibrium condition in the horizontal direction, i.e. net force acting on the body equals the rate of change of momentum of the body. From the FBD of the cart, we have

[tex]F-f-ma=0[/tex]

[tex]\Rightarrow F=f+ma[/tex]

[tex]\Rightarrow F=\mu mg+ma[/tex]

[tex]\Rightarrow F=m(\mu g +a)[/tex]

[tex]\Rightarrow F=65(0.01\times 9.81 + 0.3)[/tex]

[tex]\Rightarrow F=25.88 N.[/tex]

Hence, 25.88 N force required to accelerate the body with 0.03 [tex]m/s^2[/tex] .

Answer

No dan should reduce his discretionary apendings.

Explanation:

Answer:

Explanation:

Cardinality of a set is the number of elements in that set. Given the set.

F= {mango, apple, banana, orange)​, we are to determine the cardinality of the set i.e the amount of fruit present in the set. Cardinality of the set F is represented as n(F).

Since there are 4 different fruit in the given set F, hence the cardinality of the set F is n(F) = 4

Explanation:

At a height of 93 m, the gravitational potential energy is given by :

P = mgh

Where, m is the mass of a penny

We can find its mass.

[tex]m=\dfrac{P}{gh}\\\\m=\dfrac{3}{9.8\times 93}\\\\m=0.00329\ kg[/tex]

We need to find how much potential energy will have transformed into kinetic energy at the half way point (46.5m). It can be calculated as :

[tex]E=mgh'\\\\E=0.00329\times 9.8\times 46.5\\\\E=1.499\ J[/tex]

Hence, a penny will have transferred 1.499 J of potential energy into kinetic energy.