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3 years ago

Which of the following has no acceleration?

Physics

2 answers:

djverab [1.8K]3 years ago

7 0

D. As acceleration is a change of speed or direction.

Fantom [35]3 years ago

6 0

'Acceleration' means ANY change in speed or direction
of motion.

A). Acceleration. The satellite is in a curved orbit, so
its direction is constantly changing.

B). Acceleration. The person is on a curved track, so
his direction is changing.

C). Acceleration. The car is following a curve, so
its direction is changing.

D). No acceleration. Straight, at constant speed.
No change of speed or direction.

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Calculate the potential V(r) for r>rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.)

Marat540 [252]

Answer:

The potential for r > rb is equal to zero.

Explanation:

For r > rb, the potential is:

$V=\frac{Kq}{r}$

Then, the net potential is:

$V_{(r)} =\frac{K(+\epsilon )}{r} +\frac{K(-\epsilon )}{r}$

$K=\frac{1}{4\pi \epsilon _{o} }$

$V_{(r)} =\frac{K(+\epsilon )}{r} -\frac{K(\epsilon )}{r}\\V_{(r)}=0$

8 0

4 years ago

A uniform solid sphere of radius r = 0.510 m and mass m = 15.5 kg turns counterclockwise about a vertical axis through its cente

klasskru [66]

Answer:

5.160384 kg*m²/s

Explanation:

The vector angular momentum P can be found using the following expression:

P = I * w

I refers to the inertia, that for a sphere is found using the expression:

I = $\frac{2}{5}$ * m * r² = $\frac{2}{5}$ * 15.5kg * (0.510m)² = 1.61262 kg*m².

The angular velocity w is given by the problem, and has a value of 3.2 rad/s.

Replacing the data we get:

P = 1.61262 kg*m² * 3.2 rad/s = 5.160384 kg*m²/s

5 0

3 years ago

During a 400-m race, a runner crosses the 100 m mark with a velocity of 12 m/s. What would be her final position if she maintain

Cerrena [4.2K]

Answer:

Explanation:

Given that

Total race distance is 400m

Her initial velocity was 0m/s²

At the 100m mark, after she has travelled 100m, her final velocity was v=12m/s²

Using equation of motion

Let determine her constant acceleration

v²=u²+2as

12²=0²+2×a×100

144=0+200a

144=200a

a=144/200

a=0.72m/s²

Then we want to know her position after another 10second

So total time is 10+12=22seconds

Then, using equation of motion

Let determine his postion

S=ut+½at²

S=0•t+½×0.72×22²

S=0+174.24

S=174.24 m

Her position will be 174.24m

7 0

4 years ago

Read 2 more answers

A technician attaches one lead of a digital voltmeter to the ground terminal of the TP sensor and the other meter lead to the ne

wariber [46]

Answer:

Neither A or B

Explanation:

The 37.3mv is not the signal voltage

sensor ground circuit does not has excessive resistance.

5 0

3 years ago

An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases

andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0

3 years ago