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3 years ago

To move a heavy object like a refrigerator what could be used to help decrease the frictional force

Physics

1 answer:

butalik [34]3 years ago

3 0

Answer:

Put a lubricant between the surface of the object and the floor. ... Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily. D.

Explanation:

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two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of m

alexdok [17]

The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.

F α Qq

If the charges are now half the initial charges:

<span>F α (1/2)Q *(1/2)q
</span>
F α (1/4)Q<span>q

The new force when the charges are each halved is (1/4) the first initial force experienced at full charge.</span>

4 0

3 years ago

Read 2 more answers

Why is the answer C?

4vir4ik [10]

Explanation:

We want to find the statement that is proven by the fact that the balls reach the same height.

A isn't supported by the evidence. Balls can reach the same height without having the same initial speed.

B isn't supported by the evidence. Balls can reach the same height without having the same launch angle.

C is supported. Projectiles spend the same amount of time going up as they do coming down, so if two projectiles reach the same height, then they must spend the same amount of time in the air.

D isn't supported by the evidence. Balls thrown at the same speed and complementary angles have the same range but different heights.

E isn't supported by the evidence. The mass of the ball doesn't affect the height.

7 0

3 years ago

Drag the item from the item bank to its corresponding match.1.A fireman turns on his hose and is knocked backwards.2.It takes le

creativ13 [48]

Answer:

1 - third law

2 - second law

3 - first law

4 - third law

5 - second law

6 - first law

Explanation:

First law

In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.

Second law

In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration, a of the object

F = ma.

Third law

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

7 0

3 years ago

Describe the location of point a. explain what you used as a reference point for you your location

Gwar [14]

I used rise and run to see where point a lies .
That's kind of an answer you could you right?

3 0

3 years ago

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e =

Svetllana [295]

Answer:

Explanation:

(a) Rigel, 2.7x10^32W, T = 11,000K

But L = 4pR²sT⁴

L = 2.7x10^32W, T = 11,000K, s= 5.67 x 10^-8, R= radius in meters

Rigel parallax, p = 0.00378 arc sec

Substituting the various values and making R the subject of the formula

R² = L/(4psT⁴)

R² = 2.7x10^32/(4 x 0.003878 x 5.67x10^-8 x (11,000)⁴)

R² = 2.7x10^32/1.2877x10^7

R² = 2.096761668 x 10^25

R = 4.579041021 x 10^12meters

(b)

Procyon B, 2.1x10^23W, T = 10,000K

But L = 4pR²sT⁴

L = 2.1x10^23W, T = 10,000K, s= 5.67 x 10^-8, R= radius in meters

Procyon B parallax, p = 0.00284 arc sec

R² = 2.1x10^23/(4 x 0.00284 x 5.67x10^-8 x (10,000)⁴)

R² = 2.1x10^23/(6.441 x 10^6)

R² = 3.26036 x 10^16

R = 1.80565 x 10^8 meters

(c) The radius of Rigel is given as 4.579041021 x 10^12meters and the radius of Procyon B is given by 1.80565 x 10^8 meters shows the remarkable difference between a super-giant star(Rigel) and a white dwarf star (Procyon B)

The radius of the sun a red star is 6.96 x 10^8meters which shows a certain level of resemblance with the size of a dwarf white star Procyon B.

The sun is larger than Procyon B as estimated above.

7 0

3 years ago