Ask question

Ask question

All categories

3 years ago

8

A 40 kg boy is moving upwards upwards in a lift with an acceleration of 2 m /s ^2 what would be the weight felt by him if measuu

red in a scale?

1 answer:

neonofarm [45]3 years ago

5 0

Answer:

<h2> 48kg</h2>

$f = ma \\$

$w \: = \frac{f}{g} \\$

Explanation:

$f \: = ma \\ f - mg = ma \\ f = ma + mg \\ f = 40 \times 2 + 40 \times 10 \\ f = 480$

$w = \frac{f}{g} \\w = \frac{480}{10} \\ w = 48kg$

You might be interested in

Suppose that it takes a simple pendulum 1.2 seconds to swing from its leftmost point to its rightmost point. what is the period

Mariana [72]

The time period of the simple pendulum is 2.4 seconds.

Given the data in the question;

Time taken to swing from leftmost point to rightmost point; $t = 1.2s$

Period of the pendulum; $T = \ ?$

<h3>What is Period?</h3>

Period is the time needed for a complete cycle of vibration to pass a given point.

Period of a pendulum is the of time needed for it to complete one full back-and-forth motion. It is the time required to for the pendulum to swing from leftmost point to rightmost point and back to leftmost point.

Now, if it took the pendulum $1.2s$ to swing from leftmost point to rightmost point, it will also take the pendulum $1.2s$ to swing back to its original position( leftmost point )

Hence,

$T$ = time taken to swing from leftmost to rightmost + time taken to swing from rightmost point to leftmost point.

$T = 1.2s + 1.2s\\\\T = 2.4s$

Therefore, the time period of the simple pendulum is 2.4 seconds.

Learn more about Time Period: brainly.com/question/27135322

7 0

3 years ago

When the velocity of the car is constant the force of friction is?

miv72 [106K]

..... equal to the force from the engine pushing the car forward.

That's why the car is not accelerating. The horizontal forces on it
are balanced.

6 0

3 years ago

Which law describes the interactions between charged particles when they are not in contact?

natulia [17]

The answer would be the coulombs law.

7 0

3 years ago

Read 2 more answers

A ship leaves a port at noon and travels due west at 20 knots. At 6 PM, a second ship leaves the same port and travels northwest

Amanda [17]

Answer:

$v = 12.44 Knots$

Explanation:

First ship starts at Noon with speed 20 Knots towards West

now we know that 2nd ship starts at 6 PM with speed 15 Knots towards North West

so after time "t" of 2nd ship motion the two ships positions are given as

$r_1 = 20(t + 6)\hat i$

$r_2 = 15(t)(cos45\hat i + sin45\hat j)$

now we can find the distance between two ships as

$x = \sqrt{(20(t + 6) - 10.6 t)^2 + (10.6t)^2}$

now we have

$x^2 = (120 + 9.4 t)^2 + (10.6 t)^2$

$x^2 = 200.72 t^2 + 14400 + 2256 t$

now we will differentiate it with respect to time

$2x\frac{dx}{dt} = 401.44 t + 2256$

here we know that

$t = \frac{90}{15} = 6 hours$

so we have

$x = 187.5$

now we have

$2(187.5) v = 401.44(6) + 2256$

$v = 12.44 Knots$

5 0

3 years ago

A horizontal pull A pulls two wagons over a horizontal frictionless floor, the first wagon is 500N, the second is 2000 N. The te

MrMuchimi

<span>The force will be zero if the wagons are moving at a constant speed (i.e. not accelerating), as there is no frictional force to overcome. If the wagons are accelerating, the force will be proportional to the acceleration, and 20% of the force applied by A.</span>

5 0

4 years ago