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Vedmedyk [2.9K]

3 years ago

15

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.5

0 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

1 answer:

givi [52]3 years ago

7 0

Answer:

22.1 m

Explanation:

$v_{o}$ = initial speed of ball = 14.3 m/s

$\theta$ = Angle of launch = 27°

Consider the motion of the ball along the vertical direction.

$v_{oy}$ = initial speed of ball = $v_{o} Sin\theta = 14.3 Sin27 = 6.5 ms^{-1}$

$a_{y}$ = acceleration due to gravity = - 9.8 ms⁻²

$t$ = time of travel

$y$ = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

$y = v_{oy} t + (0.5) a_{y} t^{2} \\- 3.50 = (6.5) t + (0.5) (- 9.8) t^{2}\\- 3.50 = (6.5) t - 4.9 t^{2}\\t = 1.74 s$

Consider the motion of the ball along the horizontal direction.

$v_{ox}$ = initial speed of ball = $v_{o} Cos\theta = 14.3 Cos27 = 12.7 ms^{-1}$

$X$ = Horizontal distance traveled

$t$ = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

$X = v_{ox} t\\X = (12.7)(1.74)\\X = 22.1 m$

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The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o

blsea [12.9K]

Answer:

$F_f=840N$

Explanation:

From the question we are told that

Weight of fireman $W_f= 87.5kg$

Pole distance $D=4.10m$

Final speed is $V_f 1.75m/s$

Generally the equation for velocity is mathematically represented as

$v^2 = v_0^2 + 2 a d$

Therefore Acceleration a

$a'= v^2 / 2 d$

$a'= 0.21m/s^2$

Generally the equation for Frictional force $F_f$ is mathematically given as

$F_f=m*a$

$F_f=m*(g-0.21)$

$F_f=87.5*(9.81-0.21)$

Therefore

$F_f=840N$

6 0

3 years ago

The rear window in a car is approximately a rectangle, 2.0 m wide and 0.65 m high. The inside rearview mirror is 0.50 m from the

Ymorist [56]

Answer:

0.43 m

Explanation:

Angle of incident and angle of reflection is same.

tan Θh = L' / x (eye)

L' = Length of the window

x (eye) = Distance of the mirror from the eye

tan Θh = L / (x (eye) + xw)

xw = Distance of the mirror from the window

L'/ x (eye) = L / ( x (eye) + xw)

L' = L*x (eye) / ( x (eye) + xw)

L' = (2*0.5) / (0.5 + 1.8)

L' = 0.43 m

5 0

3 years ago

Forces that act on an object but are not equal in size

mezya [45]

Forces that are equal in size but opposite in direction and do not cause a change in an object's movement are called balanced forces.

forces that aren't equal in size and do cause a change in movement (what it seems like you're asking for) are called UNBALANCED FORCES

so answer (in case that wasn't clear, as I'm tired) : unbalanced forces

7 0

3 years ago

What is a random motion

Leokris [45]

Answer:

Random Motion is a motion in which an object didn't go in a straight manner, for ex: zig zag lines, curved, etc.

Explanation:

7 0

3 years ago

Read 2 more answers

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

Nat2105 [25]

Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

8 0

3 years ago