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Tatiana [17]
3 years ago
8

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is

Physics
1 answer:
Bumek [7]3 years ago
7 0

Answer:

Answer is 25 kg m/s.

Explanation:

Given Data ;

mass = 0.5 kg

velocity = 10 m/s

Find ;

K.E = ?

Formula ;

KE = 1/2 mv 2

Solution:

KE = 1/2(0.5)(10)²(kg)(m/s)

     =25 kg m/s

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It is D, you can not replace minerals. This makes them valuable.
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Grindstone: Rotational Dynamics and Kinematics You have a grindstone (a disk) that is 133.0 kg, has a 0.635 m radius, and is tur
Colt1911 [192]

Answer:

Ff = 839.05 N

Explanation:

We can use the equation:

Ff = μ*N

where <em>N</em> can be obtained as follows:

∑ Fc = m*ac   ⇒   N - F = m*ac = m*ω²*R    ⇒  N = F + m*ω²*R

then if

F = 32 N

m = 133 Kg

R = 0.635 m

ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s

we get

N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N

Finally

Ff = μ*N = 0.10*(8390.53 N) = 839.05 N

3 0
3 years ago
Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav
DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s

Time taken to reach 12% of light speed is 3673469.39 seconds

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m

The distance it would have to travel is 6.61×10¹⁴ m

7 0
3 years ago
Which TWO statements about the substances in the experiment are true?
nataly862011 [7]

Answer:

A and B are correct both are correct

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2 years ago
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If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is
zysi [14]

Answer:

q_2=2.47\times 10^{-4}\ C

Explanation:

The charge on one object, q_1=9.9\times 10^{-5}\ C

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C

So, the charge on the other sphere is 2.47\times 10^{-4}\ C.

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