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3 years ago

Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?

Physics

1 answer:

alexandr402 [8]3 years ago

7 0

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.

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A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

5 0

3 years ago

A thin layer of liquid methylene iodide (n = 1.756) is sandwiched between two flat, parallel plates of glass (n = 1.50). What mu

Harrizon [31]

Answer:

t = 96.1 nm

Explanation:

For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength

now we know that the path difference of two reflected light from thin liquid layer is given as

$2\mu t - \frac{\lambda}{2} = N\lambda$

here we know that

$\mu = 1.756$

t = thickness of layer

N = 0 (for minimum thickness of layer)

$\lambda = 675 nm$

now we have

$2(1.756) t = \frac{675 nm}{2}$

$t = 96.1 nm$

5 0

3 years ago

A taxi is travelling at 15m/s. Its driver accelerates with acceleration 3m/s^2 for 4 s. What would be the new velocity?..... pls

Aleksandr-060686 [28]

Answer:

27 m/s

Explanation:

Given:

v₀ = 15 m/s

a = 3 m/s²

t = 4 s

Find: v

v = at + v₀

v = (3 m/s²) (4 s) + (15 m/s)

v = 27 m/s

5 0

3 years ago

Read 2 more answers

A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme

spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

$dy = \sqrt{d^2 - dx^2}$

$dy = \sqrt{3.85^2 - 3^2} = 2.41 km$

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0

3 years ago

Which type of energy will definitely not be used in the lighting of a match?

In-s [12.5K]

I say that the answere would be B

4 0

3 years ago