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Ksivusya [100]
3 years ago
15

a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed

of the car is
Physics
1 answer:
Juliette [100K]3 years ago
8 0
  • Mass of the car (m) = 2000 Kg
  • Initial velocity (u) = 15 m/s
  • Force (F) = 10000 N
  • Time (t) = 3 s
  • Let the acceleration be a.
  • By using the formula, F = ma, we get,
  • 10000 N = 2000 Kg × a
  • or, a = 10000 N ÷ 2000 Kg
  • or, a = 5 m/s^2
  • Let the final velocity be v.
  • By using the formula, v = u + at, we get,
  • v = 15 m/s + 5 m/s^2 × 3 s
  • or, v = 15 m/s + 15 m/s
  • or, v = 30 m/s

<u>Answer</u><u>:</u>

<em><u>The </u></em><em><u>new </u></em><em><u>sp</u></em><em><u>e</u></em><em><u>ed </u></em><em><u>of </u></em><em><u>the </u></em><em><u>car </u></em><em><u>is </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>m/</u></em><em><u>s.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

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A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert
strojnjashka [21]

Answer:

v=10m/sec

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

F_N+Fg

where

F_N =forceof the normal

Fg= force due to gravity

Generally the net force is given to be FC(force towards center)

F_C =F_N + Fg

F_N =F_C -Fg

F_N=F_C-F_g

F_N=\frac{mv^2}{R} -mg

Mathematical we can now derive V

m_g + \frac{8m}{4}= \frac{mv^2}{8}

\frac{5mg}{4} =\frac{mv^2}{8}

v^2 =\frac{40*10}{4}

v=10m/sec

Therefore the speed of the roller coaster is given ton be v=10m/sec

5 0
3 years ago
A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.
DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy Q_{out}=1000\ Btu

Work done W_{cycle}=300\ Btu

We need to calculate the coefficient of performance

Using formula of  the coefficient of performance

COP=\dftrac{Q_{in}}{W_{cycle}}

We need to calculate the Q_{in}

W_{cycle}=Q_{out}-Q_{in}

Put the value into the formula

300=1000-Q_{in}

Q_{in}=300-1000

Q_{in}=700\ Btu

Now put the value of Q_{in} into the formula of COP

COP=\dfrac{700}{300}

COP=\dfrac{7}{3}=2.33

Hence, The coefficient of performance for the cycle is 2.33.

5 0
3 years ago
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

  • Mass of the first puck = m_1\ =\ 5\ kg
  • Mass of the second puck = m_2\ =\ 3\ kg
  • initial velocity of the first puck = u_1\ =\ 3\ m/s.
  • Initial velocity of the second puck = u_2\ =\ -1.5\ m/s.

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0
3 years ago
Which term describes a possible explanation of, or answer to, a scientific question that is based on prior knowledge or research
suter [353]
I think it would be called a hypothesis.
3 0
3 years ago
Read 2 more answers
How does the motion of the molecules change when you increase the temperature
borishaifa [10]

Answer:

They speed up as temperature increases.

Explanation:

As temperatures increases within a molecule the particles will speed up.

6 0
3 years ago
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