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3 years ago

What is kenetic energy of a 6 kg ball going 1.9m/s?

Physics

1 answer:

Marianna [84]3 years ago

4 0

M = 5kg
v = 1.9 m/s

KE = 1/2 mv^2 = 1/2 * 6 * 1.9^2 = 10.83 J

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The spring is compressed a total of 3.0 cm, and used to set a 500 gram cart into motion. Find the speed of the cart at the insta

BartSMP [9]

Answer:

1.15 m/s

Explanation:

Part of the question is missing. Found the missing part on google:

"1. A hanging mass of 1500 grams compresses a spring 2.0 cm. Find the spring constant in N/m."

Solution:

First of all, we need to find the spring constant. We can use Hooke's law:

$F=kx$

where

$F=mg=(1.5 kg)(9.8 m/s^2)=14.7 N$ is the force applied to the spring (the weight of the hanging mass)

x = 2.0 cm = 0.02 m is the compression of the spring

Solving for k, we find the spring constant:

$k=\frac{F}{x}=\frac{14.7}{0.02}=735 N/m$

In the second part of the problem, the spring is compressed by

x = 3.0 cm = 0.03 m

So the elastic potential energy of the spring is

$U=\frac{1}{2}kx^2=\frac{1}{2}(735)(0.03)^2=0.33 J$

This energy is entirely converted into kinetic energy of the cart, which is:

$U=K=\frac{1}{2}mv^2$

where

m = 500 g = 0.5 kg is the mass of the cart

v is its speed

Solving for v,

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.33)}{0.5}}=1.15 m/s$

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3 years ago

An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

Ivahew [28]

Answer:

The electric field is directed toward the electron and has a magnitude of $E=\frac{ke}{r^{2} }$ .

Explanation:

An electric field is define as the surrounding of charges which exert a force on each other and this force can be attractive or repulsive depends on the charge.

In the given case electron is given and the magnitude of charge on electron is $e=1.6\times 10^{-19}C$

Electric field can be represented as,

$E=\frac{kQ}{r^{2} }$

Here, r is the distance between the point ande charge, k is the electric field constant and Q is the charge.

In the given question an electron is given so electric field will be,

$E=\frac{ke}{r^{2} }$

As we know that electric field start from the positive charge and vanish in the negative charge.

So, here the electric field will be $E=\frac{ke}{r^{2} }$ and it is directed toward the electron because of negative charge on the electron.

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