Ii. Which of these has a heating element with a low melting point?

A block of mass

Gennadij [26K]

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

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6 0

2 years ago

Boat A and Boat B have the same mass. Boat A's velocity is three times greater than that of Boat B. Compared to

Zarrin [17]

Answer:

nine times as much.

Explanation:

K.E of A = 9 times K.E of B

7 0

3 years ago

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This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

Dmitry [639]

Answer:

Part a)

$F = 135.7 N$

Part b)

$F = 62.5 N$

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg + F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg + \mu(Fsin\theta)$

$F(cos\theta - \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta - \mu sin\theta}$

$F = \frac{55}{cos50 - 0.310(sin50)}$

$F = 135.7 N$

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg - F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg - \mu(Fsin\theta)$

$F(cos\theta + \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta + \mu sin\theta}$

$F = \frac{55}{cos50 + 0.310(sin50)}$

$F = 62.5 N$

6 0

3 years ago

How is water introduced into rivers and streams?

mafiozo [28]

Rain fall which is called run off

3 0

3 years ago

Read 2 more answers

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

6 0

3 years ago