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3 years ago

What is the distance traveled when a car goes 20 m/s for 60 seconds?

Physics

1 answer:

AfilCa [17]3 years ago

5 0

Answer:

1200 m

Explanation:

s= vt

s= 20m/s × 60 s

s= 1200 m

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You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa

rusak2 [61]

The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this

Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²

Height = (4.9 m/s²) (5.76 s²)

Height = (4.9/5.76) meters

Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)

Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.

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3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago

If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Cal

Alekssandra [29.7K]

The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

<h3>emf induced in the loop</h3>

The magnitude of e.m.f induced in the loop is calculated as follows;

emf = dФ/dt

Ф = 6t² + 7t

dФ/dt = 12t + 7

at t = 2 seconds

emf = dФ/dt = 12(2) + 7 = 31 V

Thus, the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

Learn more about emf here: brainly.com/question/24158806

#SPJ1

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1 year ago

Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s,

SSSSS [86.1K]

Answer:

Time period between the successive beats will be 0.1703 sec

Explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano $A\lambda _A=0.766m$

Wavelength of piano $B\lambda _B=0.776m$

So frequency of piano A $f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz$

Frequency of piano B $f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz$

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period $T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec$

So time period between the successive beats will be 0.1703 sec

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3 years ago

Describe rotational motion

ratelena [41]

Rotational motion may be described analytically for bodies undergoing pure rotation.

3 0

2 years ago