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3 years ago

What would happen if the gravity on earth decreases by a small about?

2 answers:

6 0

If the gravity decreases on the earth, we will be weightless and pressure will be less exerting on us and we will be able fly

Natali [406]3 years ago

6 0

a loss of gravity on earth would stop pulling down the air,water and earth's atmosphere

ther would be a sudden and significant loss of air pressure

the lack of any forceful gravitational pull would turn humans

Without gravity, humans and other objects would become weightless.

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Write the answer:<br>physics ... i need help

Mariana [72]

Answer:

6 gallons

Explanation:

At 30 mph, the fuel mileage is 25 mpg.

After 5 hours, the distance traveled is:

30 mi/hr × 5 hr = 150 mi

The amount of gas used is:

150 mi × (1 gal / 25 mi) = 6 gal

7 0

3 years ago

What is the slope of the line plotted below? (-2,-1) (4,2) A. 1 B. 2 C. -0.5 D. 0.5

Dmitry [639]

Answer: D. 0.5

Explanation:

The slope formula is y2-y1/x2-x1.

2-(-1)/4-(-2) = 2+1/4+2

2+1/4+2 = 3/6 = 1/2

1/2 = 0.5

The slope is 1/2, or 0.5.

8 0

3 years ago

Determine the mechanical energy of this object a 1-kg ball rolls on the ground at <br> m/s

dedylja [7]

Mechanical energy = potential energy + kinetic energy
The ball is on the ground so it has no potential energy. that's all i know.

8 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: