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MaRussiya [10]
3 years ago
11

A particle is to move in an xy plane, clockwise around the origin as seen from the positive side of the z axis. In unit-vector n

otation, what torque acts on the particle if the magnitude of its angular momentum about the origin is (a) 4.0 kg ⋅ m2/s, (b) 4.0t2 kg ⋅ m2/s, (c) 4.0 √ t kg ⋅ m 2 / s and (d) 4.0/t2 kg ⋅ m2/s?

Physics
1 answer:
slamgirl [31]3 years ago
6 0

Answer:

Answer of the part (a) Torque is zero .

Answer of the part(b) Net Torque is -8t N.m .

Answer of the part (c) is -2 (t)^-(1/2) N.m  .

Answer of the part (d) is 8(t)^-(3) N.m .

Explanation:

Explanation of all parts is in the following attachments.

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A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

K=\frac{1}{2}mv^2

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

7 0
3 years ago
After an afternoon party, a small cooler full of ice is dumped onto the hot ground and melts. If the cooler contained 5.50 kg of
zysi [14]

Answer:

Q = 4.40 \times 10^5 Cal

Explanation:

Here we know that initial temperature of ice is given as

T = 0^o C

now the latent heat of ice is given as

L = 80 Cal/g

now we also know that the mass of ice is

m = 5.50 kg

so here we know that heat required to change the phase of the ice is given as

Q = mL

Q = (5.50 \times 10^3)(80)

Q = 4.40 \times 10^5 Cal

3 0
3 years ago
If you were to drop a watermelon at velocity speed of 75 mph from 26400ft in the air how long until it would hit the ground
stellarik [79]

Answer:

Not 100% sure but if I'm right its 15 min.

Explanation: 1 mile is equal to 5380 feet, 26400 ÷ 5380 = 5, 75 ÷ 5 = 15

8 0
3 years ago
Terrance designed an experiment to show how heat can be transferred from one place to another. The steps of the experiment are s
MrRissso [65]
I think its a) heat transfer by radiation
6 0
3 years ago
Read 2 more answers
An object is moving in a straight line. At t = 0, its speed is 5.0 m/s. From t = 0 to t = 5.0 s, its acceleration is 2.5 m/s2. F
bagirrra123 [75]

Answer:

v1 = a t + v0 = 2.5 * 5 + 5 = 17.5 m/s after 5 sec

v2 = o * 6 + 17.5 = 17.5 m/s after 11 sec

S1 = V0 t + 1/2 a t^2 = 5 * 5 + 1.25 * 5^2 = 56.25 m

S2 = 17.5 * 6 = 105 m

S = 56.3 + 105 = 161 m

Vav = S / t = 161 m / 11 s = 14.6 m/s

3 0
3 years ago
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