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Zanzabum
4 years ago
7

A car is moving at a initial velocity of 20 meters per second accelerates at a rate of 1.5 meters per second squared for 4 secon

ds. What is the finally velocity

Physics
1 answer:
Brrunno [24]4 years ago
5 0
Answered using calculus.
Antidifferentiated the acceleration to get velocity. Added variable c as we do not know if there was an extra number there yet.
Knowing that when time is 0, the velocity is 20, we can substitute those numbers into the equation and find that c = 20.
Now we have full velocity equation: v = 1.5t + 20
Now we substitute 4 into t to find out the velocity after 4 seconds. This gives us the final answer of 26m/s

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What is the amount of work done when a force of 10N moves a 20kg mass of 8 meters?
Nataly_w [17]

Yo sup??

we should know that

work done=force*displacement

W=F.s

F=10 N

s=8 m

therefore

W=10*8

=80 N

Hope this helps.

3 0
3 years ago
A 0.041-kg bullet has a kinetic energy of 600j. What is the velocity of the bullet?
Paladinen [302]

171.0798 M/S

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

Was this helpful

5 0
3 years ago
How many atoms of Carbon (C) are in 3CO2
viva [34]

Answer:

in co2 there is one atom of carbon

Explanation:

8 0
3 years ago
A 15 g turntable is covered with a uniform layer of dry ice that has a mass of 9.0 g. The angular speed of the turntable and dry
pychu [463]

Answer:

1.2 rad/s

Explanation:

m1 = 15 g, m2 = 9 g, ω1 = 0.75 rad/s

Let the new angular speed is ω2 and the radius of the table be r.

The angular momentum is conserved when no external torque is applied.

I1 ω1 = I2 ω2

(m1 + m2)x r^2 x 0.75 = m1 x r^2 x ω2

(15 + 9) x 0.75 = 15 x ω2

ω2 = 1.2 rad/s

3 0
3 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

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

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

5 0
3 years ago
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