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solong [7]
3 years ago
8

An ostrich can run at a speed of 43 mi/hr. How much ground can an ostrich cover if it runs at this speed for 15 minutes? (Hint:

15 minutes = 0.25 hours)
Physics
1 answer:
Simora [160]3 years ago
4 0
..... It would possibly she eenejjsjejeej 1.4
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If a ball is launched horizontally at 40 m/s
vaieri [72.5K]

Answer:

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V= U+ at

where ,

  • v - final velocity
  • u - initial velocity
  • a-acceleration
  • t - time

v = u +at

As no force act on the ball ( we neglect air resistance here) no acceleration is seen,

So v = u = 40m/s

4 0
3 years ago
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p
Semenov [28]

Answer:

f_n=3.75N

Explanation:

From the question we are told that:

Frictional force F=0.150N

Coefficient of kinetic friction \mu=0.04

Generally the equation for Normal for is mathematically given by

 f_n=\frac{F}{\mu}

Therefore

 f_n=\frac{0.150}{0.04}

 f_n=3.75N

5 0
3 years ago
How do reflection and refraction of waves affect communication?​
romanna [79]

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

5 0
3 years ago
Read 2 more answers
How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?
fomenos

Answer:The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

Explanation:

7 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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