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iogann1982 [59]

2 years 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]2 years 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 = $P.E_f -P.E_i$

ΔP.E = 0 - $(PE)_i$

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.

Before starting to fall, the velocity of the skydiver is the same of the helicopter, which is zero. It is only, once she jumps out of the helicopter that her weight is not supported by the helicopter and so the gravitational force of the Earth attracts the skydiver and she starts to gain velocity at a rate equal to g (around 9.81 m/s²).

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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