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DENIUS [597]
2 years ago
11

a race car driver achieved a speed of 53m/s in 14 seconds after taking off from rest from the starting line. what was the averag

e acceleration during this time
Physics
2 answers:
lesya [120]2 years ago
5 0

Answer:

initial velocity= 0

time = 14 sec

final velocity= 53 m/s

a= v-u/t

= 53-0/14

= 3.78 m/s²

I hope it's correct

Veronika [31]2 years ago
3 0

Answer:

53/14

Explanation:

average acceleration = (Vfinal -Vintial)/ time taken

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What metric unit would you use to estimate the actual distance between Boston and New York?
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A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
Two dump trucks each have a mass of 1,500 kg. The distance of the dump truck
xxTIMURxx [149]

Answer:

6.00 x 10⁻⁸N

Explanation:

Given parameters:

Mass of each dump trucks  = 1500kg

Distance between them  = 50m

Unknown:

New gravitational force between them = ?

Solution:

From Newton's law of universal gravitation,

        F = \frac{G m1 m2}{r^{2} }  

F is the gravitational force

G is the universal gravitation constant

m is the mass

r is the distance

           F  = \frac{6.67 x 10^{-11} x 1500  x 1500}{50^{2} }    = 6.00 x 10⁻⁸N

4 0
3 years ago
the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i
Naily [24]

Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

P = \sigma AT^4

where,

P = Energy Radiated per Second = ?

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T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

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(b)

Now, for 90% radiator blackbody at 2000 K:

P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4

<u>P =  0.816 Watt</u>

7 0
3 years ago
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