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brilliants [131]
3 years ago
5

A block slides along a frictionless surface and onto a slab with a rough surface. The slab has mass of 4 kg and the block has ma

ss of 2 kg. What is the friction force on the small block at t = 1 second?
Physics
1 answer:
maxonik [38]3 years ago
3 0

Answer: the friction force on the small block at t equals 1 second is 2N

Explanation:

Given the data in the question;

from the slope in the graph provided, we will get the acceleration of the slab

At t = 1 seconds

Slope = acceleration = ( 1 - 0) / ( 2 - 0 ) = 1/2 = 0.5 m/s²

Force = ma = 4 × 0.5 = 2 N

so by Newton's third law

Force on block will be same which is 2N

Therefore the friction force on the small block at t equals 1 second is 2N

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dam is used to block the passage of a river and to generate electricity. Approximately 58.4 x 103 kg of water falls each second
mrs_skeptik [129]

Answer:

8.049 MW

Explanation:

The expression for gravitational potential energy is given as

Ep = mgh............. Equation 1

Ep = gravitational potential energy, m = mass of water, h = height, g = acceleration due to gravity.

Given: m = 58.4×10³ kg, h = 20.1 m, g = 9.81 m/s²

Substitute into equation 1

Ep =  58.4×10³(20.1)(9.81)

Ep = 1.6098×10⁷ J.

If one half the gravitational potential energy of the water were converted to electrical energy

Electrical energy = Ep/2

Electrical energy = (1.6098×10⁷)/2

Electrical energy = 8.049×10⁶ J

In one seconds,

The power generated = 8.049×10⁶ W

Power generated = 8.049 MW

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3 years ago
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The first law is about force or push and pull
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sladkih [1.3K]

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A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the s
tamaranim1 [39]

Answer:

(a) The speed is 7.96 m/s

(b) The direction is 76 degree from positive X axis in counter clockwise direction.  

Explanation:

Width of river = 280 m

speed of river, vR = 4.7 m/s towards east

speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north

vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j

(a) The velocity of boat with respect to ground is

\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s

(b) The direction is given  by

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7 0
3 years ago
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