Answer:
acceleration a = 3.67 m/s^2
Completed Question:
A block of mass 3 kg, which has an initialspeed of 6 m/s at timet= 0, slides on ahorizontal surface. If a constant friction force of magnitude 11 Newtons is exerted on the block by the surface, find the magnitude of the acceleration of the block. Answer in units of m/s 2
Explanation:
Force = mass × acceleration
F = ma ........1
Given that;
Force F = 11 N
mass m = 3 kg
Acceleration a can be derived from equation 1
Making a the subject of formula;
a = F/m
Substituting the given values;
a = 11N/3kg
a = 3.66666... m/s^2
acceleration a = 3.67 m/s^2
Answer:
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
Explanation:
In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.
Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.
The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.
<em>The phenomenon of Doppler effects is generally applicable to both sound and light. </em>
An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.
<em>Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc. </em>
Answer:
41.27m/s
Explanation:
According to law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the velocity after impact
Given
m1 = 0.2kg
u1 = 43.7m/s
m2 = 45.9g = 0.0459kg
u2 = 30.7m/s
Required
Velocity after impact v
Substitute the given parameters into the formula
0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v
8.74+1.409 = 0.2459v
10.149 = 0.2459v
v = 10.149/0.2459
v = 41.27m/s
Hence the speed of the golf ball immediately after impact is 41.27m/s
Answer:
Explanation:
Acceleration is the rate of change of an object with respect to time. It is the change in velocity over the time.
The car starts at 10 meters per second and increases to 20 meters per second in 2.5 seconds.
- = 20 m/s
- = 10 m/s
- t= 2.5 s
Substitute the values into the formula.
Solve the numerator.
Divide.
The acceleration of the car is <u>4 meters per second squared.</u>
the answer is 40.5 because you have to multiply the density and volume of the object to get the mass.