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2 years ago

An object is weighed at different locations on the

Physics

2 answers:

Step2247 [10]2 years ago

6 0

Answer:

Explanation:

The weight will always be different while mass is described as the stuff inside an object, and that stays the same.

Such as it weighs differently in space.

creativ13 [48]2 years ago

4 0

Answer:

C. The mass is constant while the weight varies

Explanation:

happy to help ya! :)

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Why doesn't the motor work?

exis [7]

Answer:

Explanation: its weird

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2 years ago

The diagram below shows a 5.00-kilogram block

bixtya [17]

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

Mass of the block, m = 5 kg

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

g is acceleration due to gravity = 10 m/s²

W = 5 x 10

W = 50 N (downwards)

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N (upwards)

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: brainly.com/question/14486416

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2 years ago

Read 2 more answers

A pendulum is raised to a height of 0.3m above its lowest point and released. What is the velocity of the pendulum at its lowest

enyata [817]

Answer:

v = 2,425 m / s

Explanation:

A simple pendulum has anergy stored at the highest point of the path and this energy is conserved throughout the movement.

highest point

Em₀ = U = m g y

lowest point

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

mg y = ½ m v²

v = √ 2gy

let's calculate

v = √ (2 9.8 0.3)

v = 2,425 m / s

3 0

2 years ago

Please help I wasn’t here for this lesson

katovenus [111]

Answer:

27 cm squared

Explanation:

have a good rest of ur day

8 0

2 years ago

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current density $J=2.8\times10^{7}\ A/m^2$

We need to calculate the current

Using formula of current density

$J = \dfrac{I}{A}$

$I=J\timesA$

Where, J = current density

A = area

I = current

Put the value into the formula

$I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2$

$I=1.97=2.0\ A$

Hence, The current is 2.0 A.

7 0

2 years ago