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andrezito [222]
3 years ago
11

Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x directi

on. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's x-component of velocity just after leaving the bat if the bat applies an impulse of −8.4N⋅s to the baseball? Enter your answer numerically in meters per second using two significant figures.
Physics
1 answer:
pogonyaev3 years ago
7 0

Answer:

V2 = -25.93 m/s

Explanation:

First of all, we know that;

Momentum = mass x change in velocity

Thus;

Momentum = m(v2 - v1)

Also, we know that;

Impulse = force x time

And also that;

Momentum = Impulse

From the question, m = 0.145kg and V1 = 32m/s while impulse = -8.4 N.s

Thus;

0.145(v2 - 32) = -8.4

Now,

0.145v2 - 4.64 = -8.4

0.145v2 = -8.4 + 4.64

0.145v2 = -3.76

v2 = -3.76/0.145

= -25.931 m/s and to two significant figures gives -25.93 m/s

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In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope
zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude g.

If we have the initial velocity v_o and its angle \theta, we can obtain the vertical component of the velocity v_{oy} using trigonometry:

v_{oy}=v_osin\theta

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

Finally, from the equation of horizontal motion with constant speed, we have that:

x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s}  }{9.8\frac{m}{s^{2} } } =7.49m

In words, the projectile travels 7.49m horizontally before it lands.

8 0
4 years ago
In a tug of war, two teams exerts a force of 30 N each. What is the net force acting on the rope?
Arte-miy333 [17]

The force exerted by each team is 30 N. The forces are in the direction opposite to each other.

Force exerted by team 1, F_{1}=30 N

Force exerted by team 2,F_{2}=-30 N

Net force, F=F_{1} +F_{2}

=30+(-30)=0

Therefore, net force on the rope is 0 N.

4 0
3 years ago
A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an a
sladkih [1.3K]

Answer:

d=1.29*10^{-6}m

Explanation:

From the question we are told that:

Distance of wall from CD D=1.4

Second bright fringe y_2= 0.803 m

Let

Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m

Generally the equation for Interference is mathematically given by

y=frac{n*\lambda*D}{d}

Where

d=\frac{n*\lambda*D}{y}

d=\frac{2*431 *10^{-9}m*1.4}{0.803}

d=1.29*10^{-6}m

8 0
3 years ago
The only two forces acting on a body have magnitudes of 20 N and 35 N and directions that differ by 80°. The resulting accelerat
pantera1 [17]

Answer:

b) 2.2 kg

Explanation:

Net force acting on an object is the sum of the  two forces acting on the body.

The net force is calculated using the parallelogram law of vectors.

F =\sqrt{{A^{2}} + B^{2}+2 A B cos \theta}

Here A = 20 N , B = 35 N and θ =80°

Net Force = F = 43.22 N

Acceleration = a = 20 m/s/s

Since F = ma, m = F/a = 43.22 / 20 = 2.161 kg = 2.2 kg

8 0
3 years ago
What does the slope on a velocity-time graph measure?
Lostsunrise [7]

Answer:The slope in the velocity-time graph represents the acceleration. The slope is defined as the ratio of change in y-axis to change in the x-axis. The slope is represented by the letter m and following is the general formula used for determining the slope:

Explanation:

6 0
3 years ago
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