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marin [14]
2 years ago
14

How much work must be done on a 20-kg go-cart to increase its speed from 5 m/s to 10 m/s?

Physics
2 answers:
allochka39001 [22]2 years ago
6 0

b 250 j

hope  this helps                            

Amiraneli [1.4K]2 years ago
5 0

Answer:

750 J

Explanation:

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A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
Veronika [31]

Answer:

14.98\ \text{kg m/s}

45.26^{\circ}

Explanation:

P_1 = Initial momentum of the pin = 13 kg m/s

P_i = Initial momentum of the ball = 18 kg m/s

P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

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3 years ago
When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.
DiKsa [7]
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2


Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
3 0
3 years ago
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