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ludmilkaskok [199]
4 years ago
15

19. A person pushes with 6.0 N for 4.0 seconds on a 2.0 kg object.

Physics
1 answer:
hram777 [196]4 years ago
5 0

Answer:24NS

Explanation:

Impulse=force x time

Impulse=6 x 4

Impulse=24NS

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A wire 1.0 m long experiences a magnetic force of 0.50 N due to a
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A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball
lora16 [44]

Explanation:

It is given that,

Mass of the ball, m = 0.06 kg

Initial speed of the ball, u = 50.4 m/s

Final speed of the ball, v = -37 m/s (As it returns)  

(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :

J=m(v-u)

J=0.06\times (-37-(50.4))  

J = -5.24 kg-m/s

(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 0.06\times ((-37)^2-(50.4)^2)

W = -35.1348 Joules

Hence, this is the required solution.

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Read 2 more answers
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

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Answer:

~Banana Fish~

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