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3 years ago

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A car is traveling at a speed of 15m/s. It experiences an acceleration of -3m/s^2, that lasts for 4 seconds. What is the final v

elocity of the car?

1 answer:

Gemiola [76]3 years ago

6 0

Answer:

27m/s

Explanation:

u=15m/s

a=3m/s^2

t=4s

v=u+at

v=15+3(4)

v=27m/s

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A steel wire of length 1.5 m and diameter 1 mm is joined to an aluminum wire of identical dimensions to make a composite wire of

Anarel [89]

Answer:

1.805 mm

Explanation:

Extension in the steel wire = WL_{steel}/AE_{steel}

Extension in the aluminium wire = WL_{Al}/AE_{Al}

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

we have:

W = mg

W = 5 × 9.8

W = 49 N

Area A = π/4 × (0.001)²

= 7.85398 × 10 ⁻⁷ m²

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

Total extension = 49/ 7.85398 × 10 ⁻⁷ ( (1.5/ 200×10⁹) + 1.5/ 70×10⁹))

Total extension = 0.0018048

Total extension = 1.805 mm

Thus, the total extension = the resulting change in the length of this composite wire = 1.805 mm

4 0

3 years ago

A stationary horn emits a sound with a frequency of 228 Hz. A car is moving toward the horn on a straight road with constant spe

stiks02 [169]

Answer: 26.84 m/s

Explanation:

Given

Original frequency of the horn $f_o=228\ Hz$

Apparent frequency $f'=246\ Hz$

Speed of sound is $V=340\ m/s$

Doppler frequency is

$\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)$

Where,

$v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}$

Insert values

$\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s$

Thus, the speed of the car is $26.84\ m/s$

4 0

3 years ago

Relative to the ground, what is the gravitational potential energy of 68 kg

Kisachek [45]

Answer:

Explanation:

p.E=mgh

P.E=68*10*443

p.E=301240j

8 0

2 years ago

The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is suf

kupik [55]

Answer:

(a) $I_{1}=3.2*10^{-3}W/m^{2}$

(b) $\beta =95dB$

Explanation:

Given data

Distance r₁=50 m

Distance r₂=2 m

Intensity I₂=2.0 W/m²

To find

(a) The Sound Intensity I₁

(b) The Sound Intensity level β

Solution

For (a) the Sound Intensity I₁

$\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2} }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2} }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2} }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}$

For (b) the Sound Intensity level β

The Sound Intensity level β is calculated as follow

$\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2} }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB$

8 0

3 years ago

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is

mezya [45]

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate = 150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

$h= \frac{1}{2}at^2$

$h= \frac{1}{2}4.71\times4.96^2$

h= 58.012 m

h≅ 58 m

4 0

3 years ago