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3 years ago

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Two boys push on a box. They each push to the left with a force of 50 N. At the same time, a third boy pushes on the box, to the

right, using a force of 50 N. What is the net force on the box?

1 answer:

Scrat [10]3 years ago

7 0

The net force is 50N to the left.

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A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit

JulsSmile [24]

I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ = $\frac{y*d}{D} = \frac{3.55*10^{-3}*1.25*10^{-4} }{2.25} = 1.97*10^{-7} m$
λ ~ 1.97 ×10⁻⁷m

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3 years ago

Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren

nikklg [1K]

Answer:

$4.77\ \text{A}$

Explanation:

F = Magnetic force = 4.11 N

$I_n$ = Net current

$I_2$ = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

$\theta$ = Angle between current and magnetic field = $65^{\circ}$

$l$ = Length of wires = 2.64 m

$I$ = Current in the other wire

Magnetic force is given by

$F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}$

Net current is given by

$I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}$

The current I is $4.77\ \text{A}$ .

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2 years ago

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor

joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is $\Delta U_1=222 J$

Work done on the system $W_1=-150 J$

For second step

change in internal Energy of the system is $\Delta U_2=123 J$

Work done on the system $W_2=-195 J$

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

$Q=\Delta U+W$

for first step

$Q_1=222-150=72 J$

$Q_2=123-195=-72 J$

overall heat added $=Q_1+Q_2$

$Q_{net}=72-72 =0$

For overall Process Heat added is 0 J

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3 years ago

A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box

Zielflug [23.3K]

m = mass of the box

N = normal force on the box

f = kinetic frictional force on the box

a = acceleration of the box

μ = coefficient of kinetic friction

perpendicular to incline , force equation is given as

N = mg Cos30 eq-1

kinetic frictional force is given as

f = μ N

using eq-1

f = μ mg Cos30

parallel to incline , force equation is given as

mg Sin30 - f = ma

mg Sin30 - μ mg Cos30 = ma

"m" cancel out

a = g Sin30 - μ g Cos30

inserting the values

1.20 = (9.8) Sin30 - (9.8) Cos30 μ

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3 years ago

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erastovalidia [21]

Answer:Water Only

Explanation:

Given

vessel is insulated therefore no heat can be added or removed i.e. heat exchange is zero

If hot water at $T_1$ is mixed with cold water at $T_2$ then at equilibrium vessel contains only water and final temperature of water will be between $T_1$ and $T_2$

Heat released by hot water is equal to heat gain by cold water .

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3 years ago