An object accelerates 1.3 m/s2 when a force of 3.7 newtons is applied to it. What is the mass of the object?
1 answer:
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Answer:
v₁ = 3.5 m/s
v₂ = 6.4 m/s
Explanation:
We have the following data:
m₁ = mass of trailing car = 400 kg
m₂ = mass of leading car = 400 kg
u₁ = initial speed of trailing car = 6.4 m/s
u₂ = initial speed of leading car = 3.5 m/s
v₁ = final speed of trailing car = ?
v₂ = final speed of leading car = ?
The final speed of the leading car is given by the following formula:
<u>v₂ = 6.4 m/s</u>
The final speed of the leading car is given by the following formula:
<u>v₁ = 3.5 m/s</u>
Refer to the diagram shown.
Because the surface is frictionless, the resistive for, R, is zero.
Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²
The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg
For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²
Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)
Because the surface is frictionless, the resistive for, R, is zero.
Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²
The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg
For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²
Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)
Answer:
(a)
(b)
Explanation:
<u>Electric Circuits</u>
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:
(a) The electromagnetic force of the battery is and its internal resistance is
. Knowing the equivalent resistance of the headlights is
, we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:
Solving for i
i=2.28\ A
The potential difference across the headlight bulbs is
(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is