Answer:
A person climbs a set of stairs
Explanation:
Potential energy is said to be possessed by an object due to its position. As the height from the ground level increase, the potential energy increases. It is calculated by the below formula as :
P = mgh
Out of the given options, the option that illustrates an increase in potential energy is option (b) i.e. a person climbs a set of stairs. As he steps one stair, its position from ground increases. It means its potential energy increases.
Explanation:
The test for hydrogen consists of holding a burning splint held at the open end of a test tube of gas. If the gas is hydrogen it burns with a loud “squeaky pop” which is the result of the rapid combustion of hydrogen with oxygen to produce water.
To develop this problem it is necessary to apply the concepts related to the Dopler effect.
The equation is defined by
Where
= Approaching velocities
= Receding velocities
c = Speed of sound
v = Emitter speed
And
Therefore using the values given we can find the velocity through,
Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1
Therefore the cars goes to 145.3m/s
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Answer:
a= - 0.79 m/s²
Explanation:
Given that
Speed ,u = 20 mi/h
We know that
1 mi/h= 0.44 m/s
Therefore ,u = 8.94 m/s
Distance ,s= 50 m
Lets take the acceleration of the car = a m/s²
The final speed of the car ,v = 0 m/s
We know that
v²= u² + 2 a s
Now by putting the values
0²= 8.94² + 2 x a x 50
a= - 0.79 m/s²
Therefore the acceleration will be - 0.79 m/s².
