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3 years ago

8

A race car has a mass of 710 kg. it starts from rest and travels 40.0 m in 3.0 s. the car is uniformly accelerated during the en

tire time. what is the net force exerted on it?

2 answers:

Oksanka [162]3 years ago

4 0

6300N is the answer i believe

goblinko [34]3 years ago

3 0

<span><span>6300 newtons.

If this was the appropriate answer make sure to mark as the brainliest!
-procklown</span></span>

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A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is

Soloha48 [4]

Answer: $V=5839.051m/s$

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;:

$T=1.26(10)^{4}s$ is the period of the satellite

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=5.98(10)^{24}kg$ is the mass of the Earth

$a$ is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity $V$ is given by:

$V=\sqrt{\frac{GM}{a}}$ (2)

Now, from (1) we can find $a$ , in order to substitute this value in (2):

$a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}$ (3)

$a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}$ (4)

$a=11705845.57m$ (5)

Substituting (5) in (2):

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}$ (6)

$V=5839.051m/s$ (7) This is the speed at which the satellite travels

6 0

3 years ago

Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par

taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ = 5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

$t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 = \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s$

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0

3 years ago

Whenever you push an object across a frictional surface, it starts out taking a lot of force to push to get the object moving. H

son4ous [18]

Answer:

The answer is A

Explanation:

Here's an example. A child is in school taking a test. They have made a mistake on a question, and want to erase it. The eraser is made out of a type of rubber, the rubber has friction, which means the eraser has something that's going to resist movement. Now the child has exerted enough force to get it moving, and it's moving, it won't stop unless the child stops exerting force to keep it moving. Both Newton's 1st and 3rd law explain the action of moving something on a surface with friction.

5 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

Andreas93 [3]

Answer:

$v=1.08\times 10^7\ m/s$

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, $\dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2$

Separation between the plates, $d=1.75\times 10^{-2}\ m$

An electron is released from rest, u = 0

Using equation of kinematics,

$v^2-u^2=2ad$ ..........(1)

Acceleration of the electron in electric field, $a=\dfrac{qE}{m}$ ............(2)

Electric field, $E=\dfrac{\sigma}{\epsilon_o}$ ............(3)

From equation (1), (2) and (3) :

$v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}$

v = 10840393.1799 m/s

or

$v=1.08\times 10^7\ m/s$

So, the electron is moving with a speed of $1.08\times 10^7\ m/s$ before it reaches the positive plate. Hence, this is the required solution.

3 0

3 years ago

The half life of Po-218 is three minutes. How much of a 2.0 gram sample remains after 15 minutes? Suppose you wanted to buy some

Greeley [361]

After 15 minutes :

M = M₀ (1/2)^(t/T)

M = 2 (1/2)^(15/3)

M = 0.0625 gr

In order to get 0.1 gr :

M = M₀ (1/2)^(t/T)

0.1 = M₀ (1/2)^(30/3)

0.1 = Mo / 1024

Mo = 102.4 gr

5 0

2 years ago