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2 years ago

A car with a mass of 1,500 kg is accelerating at a rate of

Physics

1 answer:

AURORKA [14]2 years ago

4 0

Answer:

The net force acting on the car is

Newtons.

Hope this helps you

Explanation:

Force is defined as the product of the mass of the body and its aaceleration,

⇒

Substituting the above given values we get,

(

1500

)

(

2.0

)

3000

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A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?

olya-2409 [2.1K]

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

3 0

2 years ago

during a baseball game you are running home and slide into home plate. However you come up short and you are tagged out. Which f

vovangra [49]

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

3 0

2 years ago

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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3 years ago

Do mirrors reflect light

777dan777 [17]

Answer:

Yes, Mirror are a surface that reflects light more perfectly than ordinary objects.

Explanation:

8 0

2 years ago

Current evidence suggests that many massive jovian planets orbit at very close orbital distances to their stars. How do we think

ivanzaharov [21]

Answer:

In the Solar system, the Jovian planets are farther from the Sun. Majority of the extrasolar Jovian planets are closer to their stars. These are known as "Hot Jupiters". From the studies, the reason for the existence of massive Jovian planets to be closer to their star is found to be the gravitational interaction of these planets with other massive planets which pushes them closer to their stars. These planets are formed beyond the frost line initially but later on migrate inwards.

4 0

3 years ago