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iogann1982 [59]

1 year ago

8

If you push a 200 kg box with a horizontal force of 1,172 N and kinetic friction resists the motion with a force of 962 N, what

is the resultant acceleration in m/s2

1 answer:

Lynna [10]1 year ago

5 0

Answer:

a = 1.05m.s²

Explanation:

Fnet = m×a

Fapplied - friction = m×a

1172 - 962 = 200 × a

210 = 200a

a = 1.05

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3 years ago

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener

Elina [12.6K]

Answer:

Explanation:

From the given information:

The initial PE $(PE)_i$ = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = $P.E_f -P.E_i$

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ΔP.E = $-P.E_i$

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

$\Delta P.E + \Delta K.E + \Delta U = 0$

$\Delta U = -\Delta P.E - \Delta K.E$

this can be re-written as:

$\Delta U =- (-\Delta P.E_i) - \Delta K.E$

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

$\Delta U =(70\%) \Delta P.E_i-0$

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7 0

3 years ago

Just as a skydiver steps out of the helicopter with no forward velocity someone who’s watching start stopwatch so the time is ze

Kobotan [32]

Answer: zero.

Justification:

The downward velocity of the sky diver just before starting to fall is zero, assuming that the helicopter is not moving but just hovering.

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3 years ago

A motorcycle is travelling at a constant velocity of 30ms. The motor is in high gear and emits a hum of 700Hz. The speed of soun

timurjin [86]

Answer:

a) $T=1.43\times 10^{-3}\ s$

b) $d=0.0429\ m$

c) $\lambda=0.4857\ m$

d) $f_o=767.7\ Hz$

Explanation:

Given:

velocity of the sound from the source, $S=340\ m.s^{-1}$

original frequency of sound from the source, $f_s=700\ Hz$

speed of the source, $v_s=30\ m.s^{-1}$

(a)

We know time period is inverse of frequency:

Mathematically:

$T=\frac{1}{f}$

$T=\frac{1}{700}$

$T=1.43\times 10^{-3}\ s$

(b)

Distance travelled by the motorcycle during one period of sound oscillation:

$d=v_s.T$

$d=30\times 1.43\times 10^{-3}$

$d=0.0429\ m$

(c)

The distance travelled by the sound during the period of one oscillation is its wavelength.

$\lambda=\frac{S}{f}$

$\lambda=\frac{340}{700}$

$\lambda=0.4857\ m$

(d)

observer frequency with respect to a stationary observer:

<u>According to the Doppler's effect:</u>

$\frac{f_o}{f_s}= \frac{S+v_o}{S-v_s}$ ...........................(1)

where:

$f_o\ \&\ v_o$ are the observed frequency and the velocity of observer respectively.

Here, observer is stationary.

$\therefore v_o=0\ m.s^{-1}$

Now, putting values in eq. (1)

$\frac{f_o}{700}= \frac{340+0}{340-30}$

$f_o=767.7\ Hz$

5 0

3 years ago