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2 years ago

What is the acceleration of a 4,000 kg car pushed with a force of 12,000 N?

Physics

2 answers:

jek_recluse [69]2 years ago

8 0

Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

Lady bird [3.3K]2 years ago

7 0

Answer:

3 m/s^2

Explanation:

The equation you have to use is F=ma because the problem is a Newton's 2nd law problem.

Our known values are:

F ( Force ) = 12,000 N

m ( mass ) = 4,000 kg

a ( acceleration ) = ?

Now we plug in the known values into the equation and solve

F=ma

12,000=4,000a

We have to divide 4,000 by both sides to isolate the a value

12,000/4,000=4,000/4,000a

The 4,000s on the right of the equation cancel.

And 12,000 divided by 4,000 equals 3

The acceleration (a) is 3 meters per second squared (m/s^2)

Next, check to make sure 3 does work by plugging it back into the equation.

12,000=4,000*3

12,000=12,000 ✔

As you can see, the acceleration will be 3 m/s^2

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A. Community B. Organism C. Population D. Individual

dimulka [17.4K]

Answer:

i think its B or D

Explanation:

its Individual or Organism

3 0

3 years ago

Read 2 more answers

Determine the index of refraction of glass that is struck by unpolarized light at 53.8 degrees and resulting in a fully polarize

nignag [31]

Answer:

The refractive index of glass, $\mu_{g} = 1.367$

Solution:

Brewster angle is the special case of incident angle that causes the reflected and refracted rays to be perpendicular to each other or that angle of incident which causes the complete polarization of the reflected ray.

To determine the refractive index of glass:

$tan\theta_{P} = \frac{\mu_{g}}{\mu_{a}}$ (1)

where

$\mu_{a}$ = refractive index of glass

$\mu_{g}$ = refractive index of glass

Now, using eqn (1)

$tan{53.8^{\circ}} = \frac{\mu_{g}}{1}$

$\mu_{g} = tan53.8^{\circ}$

$\mu_{g} = 1.367$

5 0

3 years ago

A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h

boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)

from (1), (2) and (3)

$a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g$

$a = \frac{4.5-3}{3+4.5+0.75}\times 9.8$

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

4 0

3 years ago

Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a

DaniilM [7]

Answer:

a) $k = 1343.6\,\frac{N}{m}$ , b) $l = 0.501\,m\,(50.1\,cm)$

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

$F \propto \Delta l$

In this case, the force is equal to the weight of the object:

$F = m\cdot g$

$F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 80.614\,N$

The spring constant is:

$k = \frac{F}{\Delta l}$

$k = \frac{80.614\,N}{0.408\,m-0.348\,m}$

$k = 1343.6\,\frac{N}{m}$

b) The length of the spring is:

$F = k\cdot (l-l_{o})$

$l = l_{o} + \frac{F}{k}$

$l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }$

$l = 0.501\,m\,(50.1\,cm)$

6 0

3 years ago

5. What is the velocity of a Nolan Catholic football player if he runs 200.0 m in 24.0 s?

Degger [83]

Answer:

$\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}$

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

Finding the velocity 

$\boxed{ \sf{velocity = \frac{distance \: travelled \: }{time \: taken}}}$

$\dashrightarrow{ \sf{velocity = \frac{200 \: m}{24 \: s} }}$

$\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}$

Hope I helped!

Best regards!

6 0

3 years ago