All categories

3 years ago

Can one of yall help???

Physics

2 answers:

Art [367]3 years ago

8 0

1,200 x 3.0 = 3600N

Vikki [24]3 years ago

5 0

Explanation:

given

mass (m) = 1200 kg

Acceleration (a) = 3 m/s²

Force (F) = ?

We know

F = m * a

= 1200 * 3

= 3600 Newton

The force acting on the car is 3600 Newton.

Hope it helps :)

You might be interested in

In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8

BlackZzzverrR [31]

Answer:

The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

F = m a

a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

P = I / c absorbent surface

P = F / A

F / A = I / c

F = I A / c

The area of area of a circle is

A = π r²

We replace

F = I π r² / c

Let's calculate

F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

F = 8.375 10⁻²³ N

Density is

ρ = m / V

m = ρ V

m = ρ (4/3 π r³)

m = 4500 (4/3 π (1 10⁻⁶)³)

m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

a = 4.44 10⁻⁹ m/s² absorbent surface

The correct answer is B

4 0

3 years ago

Which of the following are true about centripetal force? Check all that apply. A. Without centripetal force, an object cannot ac

Katyanochek1 [597]

B. Friction can be a centripetal force, such as when it keeps a car on the road going around a curve.

C. Gravity can be a centripetal force, such as when it pulls a satellite in its orbit.

Explanation:

The centripetal force is any force that keeps an object moving in circular motion, "pulling" the object towards the centre of the circular trajectory.

Several forces can act as centripetal force. Examples are:

- friction: when a car is going around the curve, is moving by circular motion. The force that keeps the car in circular motion is, in fact, the friction between the tires and the road.

- Gravity: when a satellite moves around the Earth, it is moving by circular motion. The force that keeps the satellite in circular motion is the gravitational attraction between the Earth and the satellite, that pushes the satellite towards the Earth.

The other two options are not correct because:

A) An object can also accelerate if there is no centripetal force (for example, a car speeding up on a straight road is accelerating, but there is no centripetal force since there is no circular motion

D) Centripetal force is not an outward force, since it pushes the object inwards (towards the centre of the trajectory).

3 0

3 years ago

Read 2 more answers

Semiconductors are materials that conduct electric current better than insulators but not as well as conductors.Semiconductors a

Lina20 [59]

Yes ! it is absolutely true.

3 0

3 years ago

Hey, I need help can someone help me out, please?

yan [13]

Explanation:

6) newton

7) f =ma = 15*15 = 225N

8) a= 100/20 = 5ms^-2

6 0

3 years ago

Read 2 more answers

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago