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2 years ago

A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.

Physics

1 answer:

Inessa [10]2 years ago

6 0

Hi there!

We can use Newton's Second Law:
$\Sigma F = ma$

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

$a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}$

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Which of the following best describes the circuit shown below?

Rudiy27

Answer:

Short circuit

Explanation:

The given figure shows a short circuit. It is defined as the circuit which allows the flow of electric current when there is no resistance. It shows a battery, bulb and connecting wires.

The wire across the bulb is connected from one terminal to another without any resistance in between them.

So, the correct option is (d) " short circuit ".

3 0

3 years ago

Read 2 more answers

Through Newton’s third law of motion, we know that if a rocket ship pushes down on the ground, the ground will push back up on t

vova2212 [387]

<span>C.
Equal Amount of Force</span>

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3 years ago

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2

Harman [31]

<span>When the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.

Given:
a = 86 m/s^2
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.7) + 0.5 (86) (1.7)^2
d = 124.27 m

vf = vi + at
vf = 0 + 86 (1.7)
vf = 146.2 m/s (velocity when the fuel is consumed completely)

Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 146.2 + (-9.8) (t)
t = 14.92 s

Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2)
d = 1090.53 m

Then, we determine the maximum altitude:
d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m</span>

5 0

3 years ago

An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e

frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

$Tsin\theta= \frac{mv^2}{r}$

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

$Tcos\theta = mg$

Matching both expressions with respect to the tension we will have to

$T = \frac{\frac{mv^2}{r}}{sin\theta}$

$T = \frac{mg}{cos\theta}$

Then we have that,

$\frac{\frac{mv^2}{r}}{sin\theta} = \frac{mg}{cos\theta}$

$\frac{mv^2}{r} = mg tan\theta$

Rearranging to find the velocity we have that

$v = \sqrt{grTan\theta}$

The value of the angle is 14.5°, the acceleration (g) is 9.8m/s^2 and the radius is

$r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}$

$r = \frac{11.1}{2}+2.41$

$r = 7.96m$

Replacing we have that

$v = \sqrt{(9.8)(7.96)tan(14.5\°)}$

$v = 4.492m/s$

Therefore the speed of each seat is 4.492m/s

6 0

3 years ago

Is it possible for the momentum of a system consisting of two carts on a low-friction track to be zero even if both carts are mo

choli [55]

Answer:

Yes, if the two carts are moving into opposite directions

Explanation:

The total momentum of the system of two carts is given by:

$p=m_1 v_1 + m_2 v_2$

where

m1, m2 are the masses of the two carts

v1, v2 are the velocities of the two carts

Let's remind that v (the velocity) is a vector, so its sign depends on the direction in which the cart is moving.

We want to know if it is possible that the total momentum of the system can be zero, so it must be:

$p=0\\m_1 v_1 + m_2 v_2 = 0\\m_1 v_1 = -m_2 v_2$

From this equation, we see that this condition can only occur if v1 and v2 have opposite signs. Opposite signs mean opposite directions: therefore, the total momentum can be zero if the two carts are moving into opposite directions.

4 0

3 years ago