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Marianna [84]
3 years ago
10

PLEASEEE HELPPPP

Physics
2 answers:
Evgen [1.6K]3 years ago
6 0

Answer:

The average car speed is  25.4 m/s

Explanation:

The definition of average speed is the total distance traveled between the total time

     Vm = dx/dt        (1)

To be able to use this equation let's calculate the total carriage distance

     Xt = X1 + X2

     X1 = 10 miles 1609  m / 1 miles = 16090 m

     X2 = 10 miles 1609 m / 1miles = 16090 m

     XT = 32180 m

Now let's calculate the total time

      v = d / t

      t = d / v

      t1 = x1 / v1

      t2 = x2 / v2

      t1 = 16090/22

      t1 = 731.36 s

      t2 = 16090/30

      t2 = 536.33 s

      t all = t1 + t2

      t all = 731.36 +536.33

      t all = 1267.69 s

Having these values ​​we use the initial equation

      Vm = d / t

      Vm = 32180 /1267.69

      Vm = 25.38 m / s

The average car speed is   25.4 m/s

solniwko [45]3 years ago
5 0

Answer:

Explanation:

average speed more than 25.0m/s.

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Answer:

200 Newtons

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A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player
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Answer:

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(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by0.5mv^{2}

Initial kinetic energy is 0.5\times 69\times 3.7^{2}=472.305 J

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305  J=-472.305  J

(b)

From fundamental kinematic equation

v^{2}=u^{2}+2as

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

s=\frac {v^{2}-u^{2}}{-2a} but a=\mu g hence

s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m

7 0
3 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
3 years ago
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7 0
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What is the resistance of a 3.5 m copper wire (Rho= 1.7x10-8 Ohm·m) that 1 point
VikaD [51]

Answer:

(D)

Explanation:

Given :

l=3.5 m

A=5.26*10^{-6} m^{2}

p=1.7*10^{-8}  ohm.m

Resistance can be calculated as :

R=p\frac{l}{A} \\R=1.7*10^{-8} \frac{3.5}{5.26*10^{-6} }

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Resistance of the wire will be 1.1×10^{-2} ohms

Option D is correct

4 0
2 years ago
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