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3 years ago

What is the Net Force acting on the box below?

Physics

1 answer:

Eddi Din [679]3 years ago

4 0

The Net Force would be 2 N to the left.

21 N is being used to push the box to the right and 23 N is used to push it left. There is a stronger force pushing the box towards the left. The different in the two numbers would give you the net force acting on the box and the direction of the arrow with the greatest force will tell you the direction.

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if solid figures have parallel bases and lateral faces how the base will be perpendicular the lateral face

Sonja [21]

Each face of a solid figure is called either a base or a lateral face. Solid figures generally have one or two bases. If it has two, these bases are parallel. If a figure has two parallel bases and lateral faces, such as in a prism, the bases will be perpendicular to the lateral faces.

6 0

3 years ago

David is driving a steady 26.0 m/s when he passes tina, who is sitting in her car at rest. tina begins to accelerate at a steady

nata0808 [166]

Relative speed of Tina with respect to David is given by

$v_r = v_t - v_d$

$v_r = 0 - 26$

$v_r = - 26 m/s$

now the acceleration of Tina with respect to David

$a_r = a_t - a_d$

$a_r = 2.80 - 0$

$a_r = 2.80 m/s^2$

now the relative displacement would be zero when Tina cross David

so now we have

$\deta x = 0 = v_r * t + \frac{1}{2} a_r t^2$

$0 = -26 * t + \frac{1}{2}*2.8*t^2$

$t = 18.6 s$

now the speed of Tina at this moment is given as

$v_f = v_i + a * t$

$v_f = 0 + 2.8 * 18.6$

$v_f = 52 m/s$

<em>so the speed will be 52 m/s</em>

5 0

3 years ago

What current flows through a 15 ohm fixed resistor when it operates on a 120-volt outlet?

gavmur [86]

Answer: current is 8.0 A

Explanation: R= U/I I = U/R = 120 V/15 Ω= 8.0 A

7 0

3 years ago

Suppose the sun were to suddenly disappear. what would happen to the orbital path of earth? it would stay the same, but the eart

Vilka [71]

C: It would follow a path perpendicular to the radius of its current orbit

When there is nothing to keep it going in circles, it will stop moving in circles around the sun and continue in a straight path. Think of it as a catapult. If the stick stops, the rock keeps flying straight at the speed and direction it was flung.

7 0

3 years ago

Read 2 more answers

On earth, you swing a simple pendulum in simple harmonic motion with a period of 1.6 seconds. what is the period of this same pe

polet [3.4K]

The period of a simple pendulum is given by
$T= 2 \pi \sqrt{ \frac{L}{g} }$
where
L is the pendulum length
g is the acceleration of gravity

If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
$T_e= 2 \pi \sqrt{ \frac{L}{g_e} }$
where $g_e$ is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
$T_m= 2 \pi \sqrt{ \frac{L}{g_m} }$
where $g_m$ is the acceleration of gravity on the Moon.

If we do the ratio of the two periods, we get
$\frac{T_m}{T_e} = \sqrt{ \frac{g_e}{g_m} }$
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write $g_e = 6 g_m$ and we can rewrite the previous ratio as
$\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }= \sqrt{6}$

so the period of the pendulum on the Moon is
$T_m = \sqrt{6} T_e = \sqrt{6} (1.6 s)=3.9 s$

8 0

4 years ago