Answer:
53,130 J
Explanation:
When a certain substance absorbs heat, the temperature of the substance increases according to the equation:
where
Q is the amount of heat absorbed
m is the mass of the substance
C the specific heat capacity
the change in temperature of the substance
In this problem:
m = 2.3 kg is the mass of copper
is the increase in temperature
is the specific heat of copper
So, the amount of heat absorbed is:
Answer:
557.7Nm
Explanation:
We are given that
Radius,r=1.9 m
Angular speed,
Time,t=3.5 s
Initial angular velocity,
We have to find the large torque would have to be exerted .
Where
Substitute the values
Answer:
The maximum height of the stadium is 151.3 feet
Explanation:
Please see the attached figure for a graphical description of the problem
The equation that describes a parabolic motion is:
r = (x0 + v0 t cos θ ; y0 + v0 t sin θ + 1/2 g t²)
where:
r = position vector
x0 = initial horizontal position
v0 = initial speed
θ = launching angle
y0 = initial vertical position
t = time
g = vertical acceleration due to gravity.
First, let´s find how much time the ball travels to reach a horizontal displacement of 565 feet.
The module of the rx vector (see figure) is the distance from home to the back wall of the stadium:
rx =(x0 + v0 t cos θ ; 0)
module of rx = = 565 feet
565 feet = x0 + v0 t cos θ (considering home as initial position, x0 = 0)
565 feet / v0 cos θ = t
565 feet / 176 feet/s cos 35° = t
t = 3.92 s
With this new data, we can now calculate the module of the y-component of the vector r at time t = 3.92, which is the time at which the ball travels 565 feet in the horizontal.
The y-component of vector r is:
ry =(0 ; y0 + v0 t sin θ + 1/2 g t²)
module of ry = = back wall height (h)
h = y0 + v0 t sin θ + 1/2 g t²
Replacing with the data:
h = 3 ft + 176 ft/s * 3.92 s* sin 35° + 1/2 * (-32.2 ft/s²) * (3.92 s)²
<u>h = 151.3 ft</u>
Answer:
This is the correct question.
Apply the junction rule to the junction labeled with the number1 (at the bottom of the resistor of resistance R_2).
Answer in terms of given quantities,together with the meter readings I_1 and I_2 and the current I_3.
b) Apply the loop rule to loop 2 (the smaller loop on the right).Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal.
Express the voltage drops in terms ofV_b, I_2, I_3, the given resistances, and any other givenquantities.
c) Now apply the loop rule to loop 1(the larger loop spanning the entire circuit). Sum the voltagechanges across each circuit element around this loop going in thedirection of the arrow.
Express the voltage drops in terms ofV_b, I_1, I_3, the given resistances, and any other givenquantities.
Explanation:
a. Junction rule: Kirchhoff current law (KCL)
Then, total current entering a junction equals to current leaving the junction.
Therefore,
i1=i2+i3
b. Apply Kirchoff voltage law ( KVL) to loop 2
Then, sum of voltage in the loop equals to zero.
-i3R3+i2R2=0.
Then,
i2R2=i3R3
c. Apply KVL to the loop 1
-Vb+i1R1+i3R3
Therefore,
Vb=i1R1+i3R3.
The circuit diagram is in the attachment
