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Korvikt [17]
3 years ago
7

How much kinetic energy does a 1500 kg car have if it’s moving at 15 m/s

Physics
1 answer:
Tresset [83]3 years ago
6 0

Answer:

Mass of car m = 1500 kg

velocity v = 15 m/s

kinetic energy = ½ mv2

= ½ x 1500 x (15)2    

= 1687501   kg m2 /s2

  = 168750J

Explanation:

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A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

    (v^2)/(R) = g×μ

               μ = (v^2)/(R×g)

                  =  ((25)^2)/((100)×(9.8))

                  = 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0
3 years ago
A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl
zysi [14]

Answer:\alpha =9.302\ rad/s^2

Explanation:

Given

mass of sphere m=250\ gm

diameter of sphere d=4.30\ cm

radius r=\frac{4.30}{2}\ cm

f=0.0200\ N

friction will provide resisting torque so

f\times r=I\times \alpha

where I=\text{moment of Inertia}

f=\text{friction force}

\alpha =\text{angular acceleration}

I=\frac{2}{5}mr^2

0.02\times r=\frac{2}{5}mr^2\times \alpha

\alpha =\frac{5}{2r}\times f

\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02

\alpha =9.302\ rad/s^2

(b)time taken to decrease its rotational speed by 21\ rad/s

t=\dfrac{\Delta \omega }{\alpha }

t=\dfrac{21}{9.302}

t=2.25\ s

6 0
3 years ago
Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if
sergij07 [2.7K]

Explanation:

a) I=\displaystyle \sum_{i}m_ir_i^2

where r_i is the distance of the mass m_i from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2

7 0
3 years ago
What would be the velocity of a sound wave in the air with a frequency of 260Hz?
enyata [817]

The speed of sound in air doesn't depend on the frequency of the sound.  It depends on the temperature, pressure, and humidity of the air.  It's usually around <em>340 m/s</em> .

7 0
3 years ago
How many stars are in ursa minor?
Hunter-Best [27]

Answer:

7

Explanation:

3 0
3 years ago
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