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vredina [299]
1 year ago
12

An object on Earth weighs 150 N. What is its mass?

Physics
1 answer:
Alex777 [14]1 year ago
3 0

Answer:

\tt \: mass \:  =  \frac{weight}{acceleration \: due \: to \: gravity}

\longrightarrow \tt \:  \frac{150}{10}

\longrightarrow \boxed{ \tt{15 \: kg}}

  • Our final answer is 15 kg .

----------- HappY LearninG<3 ------------

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Five 60 ohm resistors are connected in parallel. What is their equivalent resistance?
Mnenie [13.5K]

Answer:

i think the answer is 12 ohms

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7 0
3 years ago
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If you increase the amount of spinach you eat, then you will increase the iron in your blood. What's the dependent variable?
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3 0
2 years ago
Four forces are exerted on a disk of radius R that is free to spin about its center, as shown above. The magnitudes are proporti
Dmitry_Shevchenko [17]

The given magnitude of forces of F₁ = F₄, F₂ = F₃, F₁ = 2·F₂, give the

forces that exert zero net torque on the disk as the options;

(B) F₂

(D) F₄

<h3>How can the net torque on the disk be calculated?</h3>

The given parameters are;

F₁ = F₄

F₂ = F₃

F₁ = 2·F₂

Therefore;

F₄ = 2·F₂

In vector form, we have;

\vec{F_4} = \mathbf{\frac{\sqrt{3} }{2} \cdot F_4 \cdot \hat i -  0.5 \cdot F_4 \hat j}

\vec{F_2} = \mathbf{ -F_2 \,  \hat j}

Clockwise moment due to F₄, M₁ = -0.5 \times F_4 \,  \hat j  \times \dfrac{R}{2}

Therefore;

M_1  =- 0.5 \times 2 \times  F_2 \,  \hat j  \times \dfrac{R}{2} =   \mathbf{ -F_2 \,  \hat j  \times \dfrac{R}{2}}

Counterclockwise moment due to F₂ = -F_2 \,  \hat j  \times \dfrac{R}{2}

Given that the clockwise moment due to F₄ = The counterclockwise moment due to F₂, we have;

Two forces that combine to exert zero net torque on the disk are;

F₂, and F₄

Which are the options; (B) F₂, and (D) F₄

Learn more about the resolution of vectors here:

brainly.com/question/1858958

4 0
2 years ago
What is the initial position
labwork [276]

Answer:

Initial position of a body is the position of the body before accelerating or increasing its velocity the position changes and then that position is the final position.

hope it is helpful...

6 0
2 years ago
A​ right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 ​lb/ft3
tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ ft^{3} to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = \pi r^{2} =  \pi 4^{2} =16\pi ft^{2}

Now,

Work done required = \int_{0}^{8} 52\times 16\pi (21 - y)dy

                                  = 832\pi \int_{0}^{8} (21 - y)dy

                                  = 832\pi([ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\)

                                  = 113152\pi = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0
3 years ago
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