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3 years ago

Is an object moving in a circle at a constant speed accelerating?

Physics

1 answer:

hichkok12 [17]3 years ago

4 0

Answer:

yes.

Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed.

Explanation:

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A car with a velocity of 6.4 m/s, forward, accelerates to a velocity of 10.6 m/s, forward, in 16 s. What is the card acceleratio

tatuchka [14]

Answer:

acceleration = 0.2625 m/s²

Explanation:

acceleration = ( final velocity - initial velocity ) / time

Here the final velocity is 10.6 m/s and initial velocity is 6.4 m/s and time is 16 s.

using the equation:

acceleration = ( 10.6 - 6.4 ) / 16

= 0.2625 m/s²

6 0

2 years ago

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Use the mass and density data to calculate the volume of corn syrup to the nearest tenth.

Nataly [62]

41.5 is the answer that i got. hope this helps!

4 0

3 years ago

Read 2 more answers

What change will always result in an increase in gravitational force between two objects

tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.

Since the formula is

$F = G\frac{m_1m_2}{d^2}$

If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.

So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.

So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

6 0

3 years ago

A telephone line has a signal-to-noise ratio of 1000 and a bandwidth of 4 KHz. What is the maximum data rate supported by this l

ivolga24 [154]

Answer:

The maximum data rate supported by this line is 39900 bps

Explanation:

The maximum data rate supported by this line can be obtained using the formula below

c = W*log2(S/N+1)

where;

c is the maximum data rate supported by the line

W is the bandwidth = 4kHz

S/N+1 is the signal to noise ratio = 1001

c = 4*log2(1001)

c = 39868.9 ≅ 39900 bps

Therefore, the maximum data rate supported by this line is 39900 bps

5 0

3 years ago

2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

Alex777 [14]

The resultant speed of the plane is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

$ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}$

$ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}$

Hence, we have that the resultant speed of the plane is (3) 226 m/s

Have a nice day!

5 0

3 years ago