Answer:
Apparent frequency of the bell to the observer is 546.12 Hz
Explanation:
The frequency of train bell (frequency of source) = 505 Hz
The speed of train (observer) = 27.6 m/s
The speed of sound in the air is (velocity of sound) = 339 m/s
The apparent frequency of the bell to the observer is calculated as follows:
Apparent frequency of bell to the observer.
area of 0.1 m², how much force is needed to raise the customer?
STEP 1: List the known
and unknown values F =
A=
A,
STEP 2: Write the
correct equation
STEP 3: Insert the
known values into the
equation to solve for
the unknown value
Answer:
15N
Explanation:
F¹=150N
A=0.01m2²
F2=?
A2=0.1m²
P=F/A
F1/A2=F2/A1
150/0.1=F2/0.01
a.rubber
b.iron
c.copper
d.aluminum
Answer:
Work = 1167.54 J
Explanation:
The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:
Gain in K.E = Loss in P.E + Work
(0.5)(m)(Vf² - Vi²) - mgh = Work
where,
m = mass of boy = 60 kg
Vf = Final Speed = 8.5 m/s
Vi = Initial Speed = 1.6 m/s
g = 9.8 m/s²
h = height drop = 1.57 m
Therefore,
(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work
Work = 2090.7 J - 923.16 J
Work = 1167.54 J
Explanation:
15.556 metres per second
Answer:
None of the above
It should be position is changing and acceleration is constant.
Explanation:
Since the velocity is changing, this means the object is moving, so the position must also be changing.
Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the acceleration must be constant too.
So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.
So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"
In straight line motion, if velocity changes at a constant rate, then the position is changing and the acceleration is constant and non-zero. This is defined under the principles of kinematics and implies that as the velocity alters constantly, the object is in motion, hence its position is changing.
In straight line motion, if the velocity of an object is changing at a constant rate, then its position is changing and its acceleration is constant and non-zero. This condition is defined under the laws of physics, more specifically, under the study of kinematics.
The acceleration is constant because you're considering a situation where velocity is changing at a constant rate. In this case, the change in velocity is the acceleration, which is a constant and not zero. This situation is described by the kinematic equations for constant acceleration.
The position is changing because the object is moving. A change in position over time constitutes motion, and in this case, because the velocity (the rate of change of position) is changing, the object's position cannot be constant.
#SPJ3
Answer:
a = 5.53 g , a = -15g
Explanation:
This is an exercise in kinematics.
a) Let's look for the acceleration
as part of rest v₀ = 0
v = v₀ + a t
a = v / t
a = 282 / 5.2
a = 54.23 m / s²
in relation to the acceleration of gravity
a / g = 54.23 / 9.8
a = 5.53 g
b) let's look at the acceleration to stop
va = 0
0 = v₀ -2 a y
a = vi / y
a = 282/2 1
a = 141 m /s²
a / G = 141 / 9.8
a = -15g