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3 years ago

What kind of force do you exert on an object when you pull it towards you?

Physics

2 answers:

vlabodo [156]3 years ago

8 0

By definition we have to:

Applied force: It is the external force that acts directly on a body.

Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.

This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.

Answer:

you exert an Applied Force on an object when you pull it towards you

A. Applied Force

Aloiza [94]3 years ago

5 0

Answer: A, Applied force

Explanation: It is the external force that acts directly on a body. Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.

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_____ you remember to put the lid back on the jar of mayonnaise

Lilit [14]

Answer:

Did you remember to put the lid back on the jar of mayonnaise?

Explanation: Hope this helps :)

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3 years ago

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$

Generally, the equation for Rate of flow of Liquid is mathematically given as

$\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}$

Where dP is pressure difference r is the radius

$\mu$ is the viscosity of water

L is the length of the pipe

$Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}$

$Q=5.2 \mathrm{~mL} / \mathrm{s}$

In $30s the quantity that flows out of the tube

$&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}$

In conclusion, the quantity that flows out of the tube

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

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2 years ago

The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up

Brrunno [24]

Answer:

Explanation:

We shall apply concept of impulse to solve the problem .

Impulse = force x time

impulse = change in momentum

force x time = change in momentum

initial speed u = 24 km/h = 6.67 m /s

final speed v = 65 km/h = 18.05 m /s

change in momentum = m v - mu

= m ( v-u )

= 1350 ( 18.05 - 6.67 )

= 15363 kg m/s

F x 18 = 15363

F = 853.5 N .

4 0

4 years ago

A. If a T-Rex runs 20 meters in 4 seconds, what is it’s average speed?

mariarad [96]

5m/s

100m

Explanation:

Average speed is sum of distance distance traveled in a given time by a body.

Average speed= $\frac{total distance covered}{time taken}$

Distance = 20m

time = 4s

Average speed = $\frac{20}{4}$ = 5m/s

For the spaceship;

Distance covered = speed x time

Speed = 50m/s

time = 2s

Distance covered = 50 x 2 = 100m

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3 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.8 x 1023 kg, a radius of 2.7 x 106 m, and no atmosphere. a 10 kg space probe is to

Tom [10]

To answer these questions just use the equations for potential energy using the mass and heights described. the potential energy at the prescribed heights = the initial kinetic energy required to reach that height.

Make sure you calculate the force of gravity on the surface using the radius of the planet.

5 0

3 years ago