Answer:
a) W = - 318.26 J, b) W = 0
, c) W = 318.275 J
, d) W = 318.275 J
, e) W = 0
Explanation:
The work is defined by
W = F .ds = F ds cos θ
Bold indicate vectors
We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces
Let's use trigonometry to break down weight
sin θ = Wₓ / W
Wₓ = W sin 60
cos θ = Wy / W
Wy = W cos 60
X axis
How the body is going at constant speed
fr - Wₓ = 0
fr = mg sin 60
fr = 15 9.8 sin 60
fr = 127.31 N
Y Axis
N - Wy = 0
N = mg cos 60
N = 15 9.8 cos 60
N = 73.5 N
Let's calculate the different jobs
a) The work of the force of gravity is
W = mg L cos θ
Where the angles are between the weight and the displacement is
θ = 60 + 90 = 150
W = 15 9.8 2.50 cos 150
W = - 318.26 J
b) The work of the normal force
From Newton's equations
N = Wy = W cos 60
N = mg cos 60
W = N L cos 90
W = 0
c) The work of the friction force
W = fr L cos 0
W = 127.31 2.50
W = 318.275 J
d) as the body is going at constant speed the force of the tape is equal to the force of friction
W = F L cos 0
W = 127.31 2.50
W = 318.275 J
e) the net force
F ’= fr - Wx = 0
W = F ’L cos 0
W = 0
Answer:
<h3>The answer is 4.29 m/s²</h3>
Explanation:
The acceleration of an object given it's velocity and time taken acting on it can be found by using the formula

From the question
velocity = 30 m/s
time = 7 s
We have

We have the final answer as
<h3>4.29 m/s²</h3>
Hope this helps you
Answer:
The correct answer is c. When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL
Explanation:
Let's analyze the situation of the globe
When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.
We can relate this force of Hooke with an elastic energy
This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.
The correct answer is c
Answer:
a. 8.33 x 10 ⁻⁶ Pa
b. 8.19 x 10 ⁻¹¹ atm
c. 1.65 x 10 ⁻¹⁰ atm
d. 2.778 x 10 ⁻¹⁴ kg / m²
Explanation:
Given:
a.
I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s
P rad = I / us
P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s
P rad = 8.33 x 10 ⁻⁶ Pa
b.
P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]
P rad = 8.19 x 10 ⁻¹¹ atm
c.
P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]
P rad = 1.67 x 10 ⁻⁵ Pa
P₁ = 1.013 x 10 ⁵ Pa /atm
P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm
d.
P rad = I / us
ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²
ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²
Answer:
18 W
Explanation:
Applying,
P = V²/R.................. Equation 1
Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs
Since: It is a series circuit,
Then,
R = R1+R2............. Equation 2
Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb
Given: R1 = R2 = 8 Ω
Substitute into equation 1
R = 8+8
R = 16 Ω
Also Given: V = 12 V
Substitute into equation 1
P = 12²/8
P = 144/8
P = 18 W