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Sedbober [7]
3 years ago
11

A tennis ball was thrown into the air with a speed of 22.5 m/s. It was caught at the same height above the ground. a. How fast w

as the ball moving at the peak of its path? b. What was the acceleration on the ball as the ball was moving upward? c. What was the acceleration on the ball as the ball was moving downward? d. What was the acceleration on the ball at the peak of its path? e. How much time was the ball in the air? f. How fast was the ball moving when it was caught?
Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0
A).  The ball was moving at zero speed at the peak.

b).  The acceleration of the ball was 9.8 m/s² downward, from
the time the hand threw it until the time the hand caught it.

c).  see the answer to b).

d).  see the answer to b).

e).  The ball was in the air for  2 (22.5/9.8) = 4.6 seconds.

f).  The ball was moving down at 22.5 m/s when it was caught.
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In 1-2 sentences describe how your nose cleans the air you inhale.
svp [43]

Answer: Your nose inhales O2 and goes through these little nose hairs to keep stuff that doesn't belong in your airway, it goes through your nose into your lungs taking in O2 for your blood. The other substances that you have inhaled get absorbed elsewhere and it is converted into CO2 which is what you exhale.

Explanation:

3 0
3 years ago
In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn
kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

s=ut+\dfrac{1}{2}at_{A}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times1.6\times t_{A}^2

t_{A}=\sqrt{\dfrac{30\times2}{1.6}}

t_{A}=6.12\ sec

We need to calculate the time for B

Using equation of motion

s=ut+\dfrac{1}{2}at_{B}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times2.0\times t_{B}^2

t_{B}=\sqrt{\dfrac{30\times2}{2.0}}

t_{B}=5.48\ sec

We need to calculate the time for longest nap

Using formula for difference of time

t'=t_{A}-t_{B}

t'=6.12-5.48

t'=0.64\ s

Hence, B can take 0.64 sec for the longest nap .

4 0
3 years ago
A beam of white light enters a prism of crown glass (n = 1.5) from air (n = 1.00). Once inside, the colors in the light disperse
eduard

Answer:

= 2.33

Explanation:

.According to snell's law:

n1sin i = n2sin r ,

where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,

i is angle of incidence,

r is angle of refraction.

Given that,

n1 = 1,

i = 51 degrees,

r = 19.5 degrees. ,

n2= ?

So,

1*sin 51 = n2 sin 19.5  

=> n2 = sin51 / sin19.5

 = 2.33

7 0
3 years ago
An object is placed at O ona number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to
kvv77 [185]

Answer:

You have a displacement of 5 units to the right.

Explanation:

First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.

6 0
2 years ago
Read 2 more answers
The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center
jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0
3 years ago
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