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PIT_PIT [208]
3 years ago
9

A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?

Physics
1 answer:
GarryVolchara [31]3 years ago
3 0
Answer: Δp = F*Δt = 500N*0.001s = 0.5Ns
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Given that the distance from the left end of the string to the first antinode is 27.5 cm , calculate the wavelength of the stand
ivolga24 [154]

Answer:

= 0.55 m

Explanation:

A standing wave is characterized by anti-nodes and nodes.

Antinodes are points on a standing wave at maximum amplitude, while nodes are points on the standing wave that are stationary and have zero amplitude.

The distance between two adjacent nodes or two adjacent anti-nodes is equivalent to half the wavelength.

Therefore, in this case the half wavelength is 27.5 cm.

Thus, wavelength = 27.5 × 2

                              = 55 cm

                             <u>= 0.55 m</u>

6 0
3 years ago
As an tempature decreases what happens to the rate of radiation?
Firlakuza [10]
As the temperature decreases, the rate of radiation goes down, but the radiation exists as long as the temperature is above the absolute zero, which is actually 0 Kelvin. 0 Kelvin equals -273°C or -460°F. All objects in the world radiate if above that temperature.
5 0
3 years ago
An object of mass 8kg is attached to massless string of length 2m and swum with a tangential velocity of 3 what is the tension o
Paul [167]

Answer:

36 N

Explanation:

If the object of mass, m = 8 kg is swung in a horizontal circle of radius, r = 2m = length of string with tangential velocity v = 3 m/s, the tension in the string is the centripetal force which is T = mv²/r

= 8 kg × (3 m/s)²/2 m

= 4 kg × 9 m/s²

= 36 N

6 0
3 years ago
Example of Newton’s law of universal gravitation…
insens350 [35]
The force that holds the gases in the sun. The force that causes a ball you throw in the air to come down again
4 0
3 years ago
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an
Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

7 0
3 years ago
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