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3 years ago

Does latitude has an effect on weight?

Physics

2 answers:

SVETLANKA909090 [29]3 years ago

7 0

Answer: yes it does

Explanation: Yes, you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.

lord [1]3 years ago

5 0

Answer:

I think so but i could be wrong..

Explanation:

You might be interested in

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

sweet-ann [11.9K]

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

$\Delta T$ is the change in temperature

In this problem:

m = 1 g is the mass of water

$C=1 cal/g^{\circ}C$ is the specific heat capacity of water

$\Delta T=1^{\circ}C$ is the change in temperature

So, the heat needed is

$Q=(1)(1)(1)=1 cal$

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: $m=1g$

- Specific latent heat of fusion of ice: $\lambda_f=80 cal/g$

So, the heat needed is

$Q=(1)(80)=80 cal$

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: $m=1g$

- Specific latent heat of vaporization of water: $\lambda_v=540 cal/g$

So, the heat needed is

$Q=(1)(540)=540 cal$

5 0

3 years ago

Because it can go from rest to 14.2 m/s (about 32 mi/h) in 2.1 s, one of the fastest-accelerating animals is the cheetah. If its

TEA [102]

Answer:

(a) $P_{power}=2688.53Watt$

(b) $P_{power}=3.605hp$

Explanation:

For part (a)

As we know that

$P_{power}=\frac{W_{work}}{t_{time}}\\As\\ W=1/2(mv_{f}^{2}-mv_{o}^{2} )So\\P_{power}=\frac{1/2(mv_{f}^{2}-mv_{o}^{2} )}{t}\\ P_{power}=\frac{1/2(56.0kg*(14.2m/s)^2-56.0kg*(0m/s)^2)}{2.1s} \\P_{power}=2688.53Watt$

For part (b)

The power in unit of horsepower is:

$P_{power}=(2688.53Watt)(\frac{1hp}{745.7Watt} )\\P_{power}=3.605hp$

6 0

3 years ago

A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 45. wha

andrew-mc [135]

Answer:

The answer is B. I had a question like this last week in my physics practice exam. dont have alot of time so I cant explain right now

Explanation:

8 0

3 years ago

Calculate the mass of an asteroid moving at 500 m/s moving with 125,000 kgm/s of momentum.

soldi70 [24.7K]

Answer:

M = 250kg

Explanation:

Ft = M x V

Where ft = momentum

M = mass

V = velocity

Given

Ft = 125,000 kgm/s

V = 500m/s

Therefore

125000 = M x 500

Divide both sides by 500

125000/500 = M x 500/500

250 = M

M = 250kg

4 0

3 years ago

A chef fills a 50 mL container with 43.5 of cooking oil. What is the density of the oil?

charle [14.2K]

$density= \frac{mass}{Volume} \\ d= \frac{m}{V} \\ d= \frac{43.5}{50} \\ d=0.87 \frac{u}{mL}$

"u" stands for "units" since the unit for mass in this problem is unknown.

8 0

3 years ago

Does latitude has an effect on weight?​

Does latitude has an effect on weight?