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4 years ago

A man with a weight of 550 N climbs a ladder to a height of 3.5m. How much work did he do?

Physics

1 answer:

Natasha2012 [34]4 years ago

7 0

Answer: 1925 joules

Explanation:

weight of man = 550 N

height climbed = 3.5m

Work = ?

Since, work is done when the man pulled his weight over the height; work is expressed as the product of weight and height climbed

i.e Workdone = 550N x 3.5m

= 1925 joules

Thus, the man did a work of 1925 joules while climbing

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A particle with charge q and momentum p, initially moving along the x-axis, enters a region where a uniform magnetic field* B=(B

VMariaS [17]

Answer:

Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

So, from the question, we are given that the charge = q, the momentum = p.

=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."

Momentum,P = mass × Velocity, v -----(1).

We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.

Radius,r = P × v / B0 -----------------------------(2).

Centrepetal force = q × B0 × v. ----------(3).

(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.

Equating Equation (2) and (3) gives;

P = B0 × q × r.

Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

4 0

3 years ago

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Brums [2.3K]

The car’s velocity at the end of this distance is 18.17 m/s.

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

Final velocity, V = 18.17 m/s

Therefore, the car’s velocity at the end of this distance is 18.17 m/s.

Read more: brainly.com/question/8898885

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2 years ago

Question 7 of 10 What is cos(22")? O A. 0.93 B. 0.22 C. 0.37 O D. 0.40

Goshia [24]

Answer:

Explanation:

3 0

3 years ago

Read 2 more answers

If an astronaut can throw a certain wrench 10.0 m vertically upward on earth, how high could he throw it on our moon if he gives

Jobisdone [24]

I think you forgot to include the acceleration due to gravity of astronauts. I assume that it is = 0.170 g. To get the answer we have to use the formula s = v0t – (1/2) At². Where s is the altitude, A is the acceleration of gravity, t is the time after throwing.

v = v0 –At

v = 0 at max altitude so v0 – At = 0

t = v0/A at max altitude

Using the formula above for the altitude:

s = v0t – (1/2) At²

s = v0(v0/A) – (1/2) A (v0/A)²

s = v0²/A – (1/2) v0²/A

s = (1/2) v0²/A

The earth: E = (1/2) v0²/g

The moon: M = (1/2)v0²(0.17g)

So, take the ratio of M/E = g/0.17g = 1/0.17 = 588

M = 5.88 E

He can throw the wrench 5.88 times higher on the moon

M =5.88 (10 m) = 58.8 meters that the can throw the wrench a little over on the moon.

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4 years ago

21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest

zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

$mgh=\frac{1}{2}mu^2$