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3 years ago

How far from the base of the platform does she land?

Physics

1 answer:

seropon [69]3 years ago

3 0

When Janet leaves the platform, she's moving horizontally at 1.92 m/s. We assume that there's no air resistance, and gravity has no effect on horizontal motion. There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.

She's in the air for 1.1 second before she hits the water.

Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform

(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>

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Which statement below accurately describes the mass of the solid in the diagram?

olasank [31]

Answer:

The answer to your question is letter C. 243.8 g

Explanation:

We just need to add up to the weight of the different scales and subtract the value of the empty beaker.

The sum of the scales is: 200 + 60 + 8.8 = 268.8 g

Weight of the solid = total weight - empty beaker

Weight of the solid = 268.8 - 25 = 243.8 g

8 0

3 years ago

Car A is traveling north on a straight highway and car B is traveling west on a different straight highway. Each car is approach

guapka [62]

Answer:

129 km/hr

Explanation:

Distance of Car A North of the Intersection, y=0.3km

Distance of Car B West of the Intersection, x=0.4 km

The distance z, between A and B is determined by the Pythagoras theorem

$z^2=x^2+y^2$

$z^2=0.4^2+0.3^2=0.25\\z=\sqrt{0.25}=0.5km$

Taking derivative of $z^2=x^2+y^2$

$2z\frac{dz}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}$

$\frac{dx}{dt}=90km/hr, \frac{dy}{dt}=95km/hr$

$2(0.5)\frac{dz}{dt}= 2(0.4)X90+2X0.3X95\\\frac{dz}{dt}=72+57=129$

The distance z, between the cars is changing at a rate of 129 km/hr.

5 0

4 years ago

Read 2 more answers

If the magnitude of b⃗ increases while its direction remains unchanged, how will the magnetic flux through the coil change?

Lapatulllka [165]

Flux = B * Area
If B increases, flux also increases.

4 0

3 years ago

What is this object? Explain how it works and which one of Newton’s laws this represents.

lyudmila [28]

Answer:

the object is called Newton's cradle balance ball you can search it on Google

4 0

3 years ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Assoli18 [71]

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

When the craft was stationary, the tension in the cable was 6500 N.

When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.

The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

Here, T is tension

F = 1800 N

W = 6500 N

Tension becomes :

$T=W-F\\\\T=6500-1800\\\\T=4700\ N$

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

7 0

3 years ago