Answer:
Explanation:
From the equation of Newton's laws of motion
v = u + at where v is final velocity , u is initial velocity and t is time.
150 = 0 + a x 3
a = 50 m / s ²
s = ut + 1/2 at² ; s is distance travelled
s = 50 x 3 + .5 x 50 x 3²
= 150 + 225
= 375 m .
Answer:
-6327.45 Joules
650.375 Joules
378.47166 N
Explanation:
h = Height the bear slides from = 15 m
m = Mass of bear = 43 kg
g = Acceleration due to gravity = 9.81 m/s²
v = Velocity of bear = 5.5 m/s
f = Frictional force
Potential energy is given by

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules
Kinetic energy is given by

Kinetic energy of the bear just before hitting the ground is 650.375 Joules
Change in total energy is given by

The frictional force that acts on the sliding bear is 378.47166 N
The current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.
<h3>What current flows through the bulb as well as the power of the bulb?</h3>
From ohm's law; V = I × R
Where V is the voltage, I is the current and R is the resistance.
Also, Power is expressed as; P = V × I
Where V is voltage and I is current.
Given that;
- Resistance R = 10.0 ohms
- Voltage V = 12.0V
- Current I = ?
- Power P = ?
First, we determine the current flow through the bulb.
V = I × R
12.0V = I × 10.0 ohms
I = 12.0 ÷ 10.0
I = 1.2A
Next, we determine the power of the bulb.
P = V × I
P = 12.0V × 1.2A
P = 14.4 Watts
Therefore, the current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.
Learn more about Ohm's law here: brainly.com/question/12948166
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Answer:
10.21°C
Explanation:
From the question,we are given;
- Quantity of heat = 32,000 Joules
- Mass of water = 750 g
- Specific heat capacity of water = 4.18 J/g°C
We are required to calculate the change in temperature;
- We need to know that quantity of heat is calculated by multiplying mass by specific heat then by change in temperature.
- That is;
Q = m × c × ΔT
Rearranging the formula;
ΔT = (Q ÷ (m × c))
= 32,000 J ÷ (750× 4.18 J/g°C)
= 10.21°C
Therefore, the change in temperature is 10.21°C