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kodGreya [7K]
2 years ago
8

Ifa truck starts from rest and it has acceleration of 4 m/s for 5 second

Physics
1 answer:
yan [13]2 years ago
6 0

Answer:

Distance 50m

final velocity 20ms^-1

s = ut +  \frac{1}{2} a {t}^{2}  \\  = 0 \times t +  \frac{1}{2} \times 4 \times  {5}^{2}   = 50m

v = u + at \\ v = 0 + 4 \times 5 \\ v = 20ms^{ - 1}

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A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the
tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

8 0
2 years ago
A tennis player swings her 1000 g racket with a speed of 11 m/s. She hits a 60 g tennis ball that was approaching her at a speed
shusha [124]

Answer:

- 3.72 Ns.

9.44 m/s

Explanation:

mass of racket, M = 1000 g = 1 kg

mass of ball, m = 60 g = 0.06 kg

initial velocity of racket, U = 11 m/s

initial velocity of ball, u = 18 m/s

final velocity of ball, v = - 44 m/s

Let the final velocity of the racket is V.

(a) Momentum is defined as the product of mass and velocity of the ball.

initial momentum of the ball = m x u = 0.06 x 18 = 1.08 Ns

Final momentum of the ball = m x v = 0.06 x (- 44) = - 2.64 Ns

Change in momentum of the ball = final momentum - initial momentum

                                                        = - 2.64 - 1.08 = - 3.72 Ns

Thus, the change in momentum of the ball is - 3.72 Ns.

(b) By use of conservation of momentum

initial momentum of racket and ball = final momentum of racket and ball

1 x 11 + 0.06 x 18 =  1 x V - 0.06 x 44

12.08 = V - 2.64

V = 9.44 m/s

Thus, the final velocity of the racket afetr the impact is 9.44 m/s .

3 0
3 years ago
The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of
Alenkasestr [34]

Answer:

a) a=(-2,-6,0)m/s^2, with a magnitude of 6.3m/s^2

b) \frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2, with a magnitude of 0.76Kgm/s^2

c) F=(-0.24,-0.72,0)N, with a magnitude of 0.76N

Explanation:

We have:

v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg

We can calculate each component of the acceleration using its definition a=\frac{\Delta v}{\Delta t}

a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\

The rate of change of momentum of the ball is \frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma

So for each coordinate:

\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N

<em>(N and </em>Kgm/s^2<em> are the same unit).</em>

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