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Mama L [17]
3 years ago
8

A force of 8,480 N is applied to a cart to accelerate it at a rate of 26.5 m/s2. What is the mass of the cart?

Physics
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

<h2>320</h2>

Just use F =ma where m is mass and a is acc.

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What is a real life example of something impermeable? (please help now!!) (due tomorrow!!)
Nutka1998 [239]
Impermeable is a substance that is water proofed e.g glass, aluminium e.t.c <span />
6 0
3 years ago
In a bumper car arena, two cars of equal mass are heading straight toward each other. The orange one is traveling at a speed of
marshall27 [118]

Answer: D.

Explanation:   Orange, at 3 meters per second if you calculate the net force being applied to the system.

hope this helps! ✌

4 0
3 years ago
Learning Goal: To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following gene
kaheart [24]

Answer:

so the transverse displacement is  0.089963 m

Explanation:

Given data

equation y(x,t)=Acos(kx−ωt)

speed  v = 9.00 m/s

amplitude A = 9.00 × 10^−2 m

wavelength λ   = 0.480 m

to find out

the transverse displacement

solution

we know

v = angular frequency / wave number

and

wave number = 2 \pi / λ  =  2 \pi / 0.480  = 13.0899 m^{-2}

angular frequency = v k

angular frequency = 9.00 × 13.0899

angular frequency = 117.8097 rad/sec = 118 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)=9.00 × 10^−2 cos(13.0899 x−118t)

when x =0 and and t = 0

maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))

maximum y(x,t)= 9.00 × 10^−2  m

and when x =  x = 1.59 m and t = 0.150 s

y(x,t)=9.00 × 10^−2 cos(13.0899 (1.59) −118(0.150) )

y(x,t)=9.00 × 10^−2 × (0.99959)

y(x,t) = 0.089963 m

so the transverse displacement is  0.089963 m

5 0
3 years ago
Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o
Anna35 [415]

Answer:

the correct one is 2. the equipotential lines must be closer together where the field has more intensity

Explanation:

The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.

In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by

           V = k q / r

also the electric field and the electuary potential are related

           E =  - \frac{dV}{dr}

therefore the equipotential lines must be closer together where the field has more intensity

When checking the answers, the correct one is 2

3 0
3 years ago
The diagram below shows eight different positions of the moon around Earth.
KATRIN_1 [288]

Answer:

6 and 8

Explanation:

The different positions of the moon, as seen from Earth, reflect how much light is being reflected off the moon from the sun.

Position 1 indicates the 3rd Quarter.

Position 2 indicates the Waning Gibbous.

Position 3 indicates the Full Moon.

Position 4 indicates the Waxing Gibbous.

Position 5 indicates the 1st Quarter.

Position 6 indicates the Waxing Crescent.

Position 7 indicates the New Moon.

Position 8 indicates the Waning Crescent.

You would read the diagram counterclockwise, with positions 7 to 3 as the moon increases light and positions 3 to 7 as the moon decreases light.

Therefore, we see that our 2 choices where we have a crescent is positions 6 and 8.

6 0
3 years ago
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