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stich3 [128]
3 years ago
12

An object traveling at a constant velocity has no net force. True or False

Physics
2 answers:
iogann1982 [59]3 years ago
7 0
I think the statement is true!
mixas84 [53]3 years ago
6 0
The answer is False

Hope this helps
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A ball of radius 0.220 m rolls along a horizontal table top with a constant linear speed of 3.70 m/s.?
4vir4ik [10]
<span>16.82 x 0.04 = 0.67 rad
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6 0
3 years ago
In a thunder and lightning storm there is a rule of thumb that many people follow. After seeing the lightning, count seconds to
lubasha [3.4K]

Answer:

2.837% less than actual value.

Explanation:

Based on given information let's calculate our value.

S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.

1mile = 1609.34meters.

percentage error is.

\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%

negative indicates less than actual value.

3 0
3 years ago
A car traveling along the highway needs a certain amount of force exerted on it to stop it in a certain distance. More stopping
galben [10]

Answer:

Higher mass or higher speed

Explanation:

Higher mass will require more force

F= ma       if m goes up  F   goes up   to stop in the same distance

4 0
2 years ago
It has been suggested that a heat engine could be developed that made use of the fact that the temperature several hundred meter
lesantik [10]

Answer:

efficiency = 5.4%

Explanation:

Efficiency of heat engine is given as

\eta = \frac{W}{Q_{in}}

now we will have

W = Q_1 - Q_2

so we will have

\eta = 1 - \frac{Q_2}{Q_1}

now we know that

\frac{Q_2}{Q_1} = \frac{T_2}{T_1}

so we have

\eta = 1 - \frac{T_2}{T_1}

\eta = 1 - \frac{273+6}{273+22}

\eta = 0.054

so efficiency is 5.4%

8 0
3 years ago
The voltage across a 5-uF capacitor is: v (t )equals 10 cos open parentheses 6000 t close parentheses space straight V. What is
mamaluj [8]

Answer:

- 0.3sin6000t A

Explanation:

Voltage, v = 10 cos 6000t V

Capacitance = 5-uF

Current flowing through, i(t)

i(t) = c * d/dt (V)

c = 5-uF = 5 * 10^-6 F

i(t) = (5 * 10^-6) * d/dt(10 cos 6000t)

d/dt(10 cos 6000t) = (10 * 6000) * (-sin 6000t)

Hence,

i(t) = (5*10^-6) * (10*6000) * (-sin 6000t)

i(t) = 5*10^-6 * 6*10^4 * - sin6000t

i(t) = 30 * 10^-2 * - sin6000t

i(t) = 0.3*-sin6000t

i(t) = - 0.3sin6000t Ampere

4 0
2 years ago
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