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ELEN [110]
3 years ago
9

A car traveling at a speed of 50.0 m/s encounters an emergency and comes to a complete stop. How much time will it take for the

car to stop if it decelerates at -4.0 m/s2?
Physics
1 answer:
ludmilkaskok [199]3 years ago
6 0

Answer:

t=1.25s

Explanation:

The formula is

a= (V1-V0) /t

t = (V1-V0)/a

V1=50m/s

V0= 0 m/s

a= - 4m/s2

t= (0-50)/-4 = 1.25s

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A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .
zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

F=m\times a

F=m\times (\dfrac{v-u}{t})

F=0.003\times (\dfrac{30}{0.06})

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

6 0
3 years ago
Someone please help. science &lt;3 thx<br> ill give 15 pts for it.
STALIN [3.7K]

Answer:

put these numbers in the boxes from up to down. hope this helps! :)

Explanation:

7

6

3

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8

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5 0
3 years ago
An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum
Tju [1.3M]

Answer:

The pressure is  P_2  =  4.25 \ a.t.m

Explanation:

From the question we are told that

   The initial pressure is P_1  =  11.2\ a.t.m

   The  temperature is  T_1 =   299 \ K

   

Let the first volume be  V_1 Then the final volume will be  2 V_1

 Generally for a diatomic  gas

           P_1 V_1 ^r  =  P_2  V_2  ^r

Here r is the radius of the molecules which is  mathematically represented as

    r =  \frac{C_p}{C_v}

Where C_p \  and\   C_v are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     C_p=7 \  and\   C_v=5

=>   r =  \frac{7}{5}

=>  11.2*( V_1 ^{\frac{7}{5} } ) =  P_2  *  (2 V_1 ^{\frac{7}{5} } )

=>   P_2  =  [\frac{1}{2} ]^{\frac{7}{5} } * 11.2

=>  P_2  =  4.25 \ a.t.m

8 0
3 years ago
A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s
Natalka [10]

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

KE_i = KE_f

KE_f = \frac{1}{2} mv^2

Here,

m = Mass

V = Velocity

Replacing,

KE_f = \frac{1}{2} (12000)(11)^2

KE_f = 72600J

Therefore the  final kinetic energy of the two car system is 72.6kJ

8 0
3 years ago
Why is understanding the concept of forces, friction and gravity important?
nydimaria [60]
Because it's the basis of how everything around you works
8 0
3 years ago
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