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OLEGan [10]
3 years ago
9

A motorcycle is traveling east at a rate of 56mph.it take 6.7 s for it to decrease its speed to 35mph. How far does the motorcyc

le travel in 6.7 s if it slows at a constant acceleration? Remeber 1 mile equals 5280 ft.
Physics
1 answer:
Phantasy [73]3 years ago
8 0
Initial speed = 56mph
Final speed = 35mph

Time taken = 6.7seconds...

Converting the time to hour.. Divide by 3600..
= 6.7/3600

=0.00186hour..

Acceleration = v-u/t

a = 35-56/0.00186
a = -11283.6mph²

The negative sign shows that it decelerated...

V² = u²+2as

(35)² = (56)² + 2×-11283.6×s
Where s is the distance covered within that time...

1225 = 3136 - 22567.2s

22567.2s = 3136-1225

22567.2s = 1911

S = 1911/22567.2

S = 0.08468miles...

But at the end of the question we were made to understand that 1miles = 5280ft

Therefore 0.08468miles = (0.08468×5280)ft

= 447. 11feets...

Which is approximately 447ft.....


Hope this helped.... ?
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A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway
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Answer:

<em>63.44 rad/s</em>

<em></em>

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet v_{1} = 250 m/s

final velocity of bullet v_{2} = 140 m/s

loss of kinetic energy of the bullet = \frac{1}{2}m(v^{2} _{1} - v^{2} _{2})

==> \frac{1}{2}*0.0033*(250^{2}  - 140^{2} ) = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = \frac{1}{2} mv^{2}

70.785 = \frac{1}{2}*0.25*v^{2}

566.28 = v^{2}

v= \sqrt{566.28} = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 =<em> 63.44 rad/s</em>

5 0
3 years ago
A block of ice at 0 degrees C, whose mass is initially 62 kg, slides along a horizontal surface, starting at a speed of 5.48 m/s
Kryger [21]

The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

<u>Explanation:</u>

Given,

Temperature, T = 0°C

Initial mass, Mi = 62kg

Speed, s = 5.48m/s

Distance, x = 26.8m

Friction is present.

Mass of ice melted = ?

We know,

The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice

and

            Kinetic Energy, KE = \frac{1}{2} mv^2

Therefore,    KE = \frac{62 X  5.48 X 5.48}{2}

KE = 930.94 Joules

Ice melting lateral heat is  334 kJ/kg = 334000 J/kg.

Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.

Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

4 0
3 years ago
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