All categories

2 years ago

You push on a box with 100 N of force, causing it to accelerate at 5 m/s?.

Physics

2 answers:

Xelga [282]2 years ago

4 0

<h3><u>Given</u><u>:</u><u>-</u></h3>

Force,F = 100 N

Acceleration,a = 5 m/s²

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the mass of the box .

<h3><u>Formula</u><u> </u><u>used:-</u><u> </u></h3>

$\bf \: Force = Mass \times Acceleration$

<h3><u>Solution:-</u><u> </u></h3>

$\sf \: Force = Mass × Acceleration$

★ Substituting the values in the above formula,we get:

$\sf \implies \: 100 = Mass \times 5$

$\sf \implies \: Mass = \cancel\dfrac{100}{5}$

$\sf \implies \: Mass = 20 \: kg$

Nezavi [6.7K]2 years ago

4 0

We are given with the amount of force applied and accleration of a certain body and we have to find the mass of the box.

Force applied = 100 N
Accleration = 5 m/s²

<u>According to Netwon's 2nd law,</u>

After deriving, we will get that

Force = Mass × Accleration

Plugging the given values to get mass,

➝ 100 N = Mass × 5 m/s²

➝ Mass = 100 / 5 kg

➝ Mass = 20 kg

Thus, the required mass of the box is 20 kg.

Carry On Learning $!$

You might be interested in

what would the net force be on the box in the problems shown below.( both force and direction). for all four diagrams. please e

DIA [1.3K]

Answer:
A) object moves 20 N [West] or -20 N [East]
B) object moves 6 N [South] or -6 N [North]
C) object moves 90 N [West] or -90 N [East]
D) object does not move and is at rest*

*Rest means 0

Why:

A)both forces from north and south that are pushing against the object neutralize each other. Assume that north is positive and south is negative: 20 [N] + (-20) [S] = 0
On West and east, you can see that west has a greater force. Assume that west is negative and east is positive: 50 [E] + (-70) [W] = -20 [E]

8 0

2 years ago

A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed

Dahasolnce [82]

<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>

6 0

2 years ago

Describe why using the simulation is a good method for studying projectiles. Clearly identify the error sources the simulation e

DiKsa [7]

Am not sure
Have a nice day

6 0

3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

2 years ago

An electric travel blanket uses a car's 12V supply. The current is 3A.

snow_lady [41]

For electrical devices . . .

   Power dissipated = (voltage) x (current) =

   (12 V) x (3.0 A) = 36 watts .

1 watt means 1 joule per second

   (36 joule/sec) x (60 sec/min) x (10 min) = 21,600 joules

5 0

2 years ago