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3 years ago

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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

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>

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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 $v=10m/sec$

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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.

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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,

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

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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.

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