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3 years ago

A physics student pushed a 50 kg load across the floor, accelerating it at a rate of 1.5 m/s . How much force did she apply?

Physics

1 answer:

garri49 [273]3 years ago

4 0

Explanation:

if you're trying to find the acceleration then it's change in velocity ÷ time taken

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Make a VIR chart.<br> All voltage, resistance, and current.

KiRa [710]

$\fbox{Hope This Helps You .}$

5 0

4 years ago

A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h

raketka [301]

Answer:

$\rho = 1848.03 kg m^{-3}$

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height of water = 20 cm =0.2 m

Pressure p = 1.01300*10^5 Pa

pressure at bottom

$P = P_{fluid} + P_{h_2 o}$

$P = P_{fluid} + \rho g h$

$P_{fluid} = P - \rho g h$

= 1.01300*10^5 - 1000*0.2*9.8

= 99340 Pa

$p_{fluid} = P_{atm} + \rho g h_{fluid}$ h_[fluid} = 0.307m

$99340 = 104900 + \rho *9.8*0.307$

$\rho = 1848.03 kg m^{-3}$

5 0

4 years ago

If you were a scientist trying to separate pollution molecules from the air, how might you do it? (This question appears on page

joja [24]

Using physical means such as electrostatic filters or mechanical filters :)

8 0

3 years ago

Read 2 more answers

The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the a

Shtirlitz [24]

Answer:

11250 N

Explanation:

From the question given above, the following data were obtained:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

Friction and normal force are related by the following equation:

F = μR

Where:

F is the frictional force.

μ is the coefficient of static friction.

R is the normal force.

With the above formula, we can calculate the frictional force acting on the car as follow:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

F = μR

F = 0.75 × 15000

F = 11250 N

Therefore, the frictional force acting on the car is 11250 N

3 0

3 years ago

An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.

jekas [21]

Answer:

$\varphi_1= 796.25 N m^2/C$

$\varphi_2= 0 N m^2/C$

$\varphi_3=686.1 N m^2/C$

Explanation:

From the question we are told that

Electric field of intensity $E= 3.25 kN/C$

Rectangle parameter Width $W=0.350 m$ Length $L=0.700 m$

Angle to the normal $\angle=30.5 \textdegree$

Generally the equation for Electric flux at parallel to the yz plane $\varphi_1$ is mathematically given by

$\varphi_1=EA cos theta$

$\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0$

$\varphi_1= 796.25 N m^2/C$

Generally the equation for Electric flux at parallel to xy plane $\varphi_2$ is mathematically given by

$\varphi_2=EA cos theta$

$\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90$

$\varphi_2= 0 N m^2/C$

Generally the equation for Electric flux at angle 30 to x plane $\varphi_3$ is mathematically given by

$\varphi_3=EA cos theta$

$\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5$

$\varphi_3=686.072219 N m^2/C$

$\varphi_3=686.1 N m^2/C$

7 0

3 years ago