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2 years ago

12

Convert the mass to kgs thanks

1 answer:

Natali [406]2 years ago

4 0

Memorize this and you'll be able to do ALL of these: <em>1 kg = 1,000 g</em>

So if you have some grams, divide the number by 1,000 to get kilograms.

1,000 g = 1.000 kg

500 g = 0.500 kg

100 g = 0.100 kg

50 g = 0.050 kg

20 g = 0.020 kg

10 g = 0.010 kg

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A bucket filled with water has a mass of 54 kg and is hanging from a rope that is wound around a 0.050 m radius stationary cylin

Lubov Fominskaja [6]

Answer:

The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

Explanation:

Given that,

Mass of bucket = 54 kg

Radius = 0.050 m

We need to calculate the magnitude of the torque the bucket produces around the center of the cylinder

Using formula of torque

$\tau=F\times r$

$\tau=mg\times r$

Where, m = mass

g = acceleration due to gravity

r = radius

Put the value into the formula

$\tau=54\times9.8\times0.050$

$\tau=26.46\ N-m$

Hence, The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

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2 years ago

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2 years ago

If a sound travels 343m/s through air and has a frequency of 800Hz, what would the wavelength be of the same sound traveling thr

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Answer:

0.625m

Explanation:

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2 years ago

In deep space, there is very little friction. Once they launch a probe into deep space, where there are no external forces actin

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3 years ago

Read 2 more answers

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

2 years ago