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REY [17]
2 years ago
14

A 0.050 kg toy truck moving right at 0.20 m/s collided with a toy car weighing 0.015 kg initially at rest, on a frictionless tra

ck. The truck keeps moving right at 0.15 m/s after the collision,
what is the final speed of the race car?
Physics
2 answers:
Kamila [148]2 years ago
7 0

Answer:

0.1667 m/s

Explanation:

m1V1 + m2V2 = m1V3 + m2V4

0.01 = ( 0.0075) + (0.015 * V4)

V4 = (0.01 - 0.0075) / (0.015)

V4= 0.1667

Arada [10]2 years ago
6 0

Answer:

0.167 m/s

Explanation:

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The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.
Salsk061 [2.6K]

Answer:

5.94\ \text{m/s}

1.7

0.577

Explanation:

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

\theta = Angle of slope = 30^{\circ}

v = Velocity of child at the bottom of the slide

\mu_k = Coefficient of kinetic friction

\mu_s = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}

The speed of the child at the bottom of the slide is 5.94\ \text{m/s}

Length of the slide is given by

l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}

v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}

The force energy balance of the system is given by

mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73

The coefficient of kinetic friction is 1.7.

For static friction

\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577

So, the minimum possible value for the coefficient of static friction is 0.577.

8 0
3 years ago
The thermal energy used by heater in 3 minutes is used to melt wax.Melting point of solid wax is 60.Specific heat of wax is 220j
Romashka-Z-Leto [24]

Answer:

m = 35 g

Explanation:

The specific heat of a material can be calculated by the following formula:

C = \frac{Q}{m}\\

where,

C = Specific Heat of Wax = 220 J/g

Q = Amount of Heat Supplied by the Heater = 7700 J

m = mass of wax melted = ?

Therefore,

220\ J/g = \frac{7700\ J}{m}\\\\m = \frac{7700\ J}{220\ J/g}\\

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