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kotegsom [21]
2 years ago
9

our battery has died and your friends push your vehicle so you can kick-start the engine. You and the vehicle have a combined ma

ss of 1600 kg. If your friends do 6000 J of work and one-third of that is dissipated by friction, how fast is your vehicle traveling?
Physics
1 answer:
Kay [80]2 years ago
7 0

Answer:2.23 m/s

Explanation:

Given

Mass of person and vehicle is m=1600\ kg

Total work done is W_t=6000\ J

Friction consumes one-third of the energy

The remaining two-third is consumed to increase the kinetic energy

\Rightarrow \dfrac{2}{3}\times 6000=\dfrac{1}{2}\times 1600\times v^2\\\\\Rightarrow 4000\times 2=1600\times v^2\\\\\Rightarrow v=\sqrt{5}\ \approx 2.23\ m/s

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What is the mass of a dog running at a speed of 5 m/s and a momentum of 120.5 kgm/s?
viktelen [127]

Given:

Momentum of the dog (p) = 120.5 kg m/s

Speed of the dog (v) = 5 m/s

To Find:

Mass of the dog (m)

Concept/Theory:

\underline{\underline{ \bf{\Large{Momentum}}}}

  • It is defined as the quantity of motion contained in a body.
  • It is measured as the product of mass of the body and it's speed.
  • It is represented by p.
  • It's SI unit is kg m/s
  • Mathematical Representation/Equation of Momentum: \boxed{ \bf{p = mv}}

Answer:

By using equation of momentum, we get:

\rm \longrightarrow m =  \dfrac{p}{v}  \\  \\  \rm \longrightarrow m =  \dfrac{120.5}{5}  \\  \\  \rm \longrightarrow m = 24.1 \: kg

\therefore Mass of the dog (m) = 24.1 kg

6 0
2 years ago
Can you help me doing an essay about actual self​
Rudik [331]

Answer:

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Explanation:

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4 0
2 years ago
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
An object moving at 30 m/s takes 5 sec. to come to a stop.  What is the object’s acceleration​
Klio2033 [76]

Answer:

6 m/s²

Explanation:

a=v/t

a= 30 m/s÷ 5 sec= 6 m/s²

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Helppp pls you have to label a-d
Mama L [17]

Answer:

A

C

D

B

Explanation.

At point A The body is at rest so k.E is zero but the height is maximum so that p.E is max.

6 0
2 years ago
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