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alina1380 [7]
3 years ago
7

With good tires and breaks, a car traveling 50 mi/hr on dry pavement can travel 400 ft when the driver reacts to something he se

es and brings the vehicle to a stop. If this is done uniformly, what is the car's acceleration?
Physics
1 answer:
Bogdan [553]3 years ago
3 0

Answer:6.72 m/s^2

Explanation:

Given

initial velocity u=50 mi/hr\approx 73.33 ft/s

Distance traveled before stopping s=400 ft

using equation of

v^2-u^2=2as

where v=final velocity

u=initial Velocity

a=acceleration

s=displacement

v=0 as car stops after travelling 400 ft

0-73.33^2=2\times a\times 400

a=-6.72 m/s^2

negative sign indicates it is deceleration

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A student performs a lab measuring the velocities of toy cars of different masses
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The answer is Car 1 and Car 2.
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A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet
Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

3 0
3 years ago
A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward
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Answer:

m=18000kg

Explanation:

From the question we are told that:

Crane Length l=20m

Crane Mass m_a=3000kg

Arm extension at lifting end l_l=15m

Arm extension at counter weight end l_c=5m

Load M_l=5000kg

Generally the equation for Torque Balance is mathematically given by

T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0

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7 0
3 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

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        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

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       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
3 years ago
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