PHYSICS COLLEGE

If i throw up an object up at 31 m/s, how long will it take to get to its highest point

Answers

Answer 1

Answer:

Answer:

Explanation:

vf=vi+at

vf=31 m/s

vi=0 m/s

a=g=9.8 m/s2

t=?

vf-vi=at

vf-vi/a=t

t=vf-vi/a

t=31 m/s-0/9.8

t=3.16 s

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What is the revolution ratio of two spur gears with one having 12 teeth and the other having 36 teeth

Answers

3:1 is the ratio i hope this belps

Consider a vertical elevator whose cabin has a total mass of 800 kg when fully loaded and 150 kg when empty. The weight of the elevator cabin is partially balanced by a 400-kg counterweight that is connected to the top of the cabin by cables that pass through a pulley located on top of the elevator well. Neglecting the weight of the cables and assuming the guide rails and the pulleys to be frictionless, determine (a) the power required while the fully loaded cabin is rising at a constant speed of 1.2 m/s and (b) the power required while the empty cabin is descending at a constant speed of 1.2 m/s. What would your answer be to (a) if no counterweight were used? What would your answer be to (b) if a friction force of 800 N has developed between the cabin and the guide rails?

Answers

Answer:

Part a)

$P = 4.71 * 10^3 Watt$

Part b)

$P = 2.94 * 10^3 W$

Part c)

$P = 9.4 * 10^3 W$

Part d)

$P = 3.9 * 10^3 W$

Explanation:

Part a)

When cabin is fully loaded and it is carried upwards at constant speed

then we will have

net tension force in the rope = mg

$T = (800)(9.81)$

$T = 7848 N$

now it is partially counterbalanced by 400 kg weight

so net extra force required

$F = 7848 - (400 * 9.81)$

$F = 3924 N$

now power required is given as

$P = Fv$

$P = 3924 (1.2)$

$P = 4.71 * 10^3 Watt$

Part b)

When empty cabin is descending down with constant speed

so in that case the force balance is given as

$F + (150 * 9.8) = (400 * 9.8)$

$F = 2450 N$

now power required is

$P = F.v$

$P = (2450)(1.2)$

$P = 2.94 * 10^3 W$

Part c)

If no counter weight is used here then for part a)

$F = 7848 N$

now power required is

$P = F.v$

$P = 7848 (1.2)$

$P = 9.4 * 10^3 W$

Part d)

Now in part b) if friction force of 800 N act in opposite direction

then we have

$F + (150 * 9.8) = 800 +(400 * 9.8)$

$F = 3250 N$

now power is

$P = (3250)(1.2)$

$P = 3.9 * 10^3 W$

A boat that travels 3.00 m/s relative to the water is crossing a river that is 1.00 km wide. The destination on the far side of the river is 0.500 km downstream from the starting point. (a) If the river current is 2.00 m/s, in what direction should the boat be pointed in order to reach the destination? (b) How much time will the trip take?

Answers

Answer:

a) 10.29° upstream

b) t=338.7s

Explanation:

If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:

$\alpha =atan((0.5)/(1))=26.56°$

$D=√(1^2+0.5^2)=1.118km$

The realitve velocity of the boat respect to the water is:

$V_(B/W)=[3*cos\beta ,3*sin\beta ]$ where β is the angle it has to be pointed at.

From the relative mvement equations:

$V_(B/W)=V_B-V_W$ where $V_B=[V*cos\alpha ,-V*sin\alpha ]$

From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:

$(3*cos\beta )^2+(3*sin\beta )^2=(V*cos\alpha )^2+(-V*sin\alpha +2)^2$

$V^2-4*V*sin\alpha -5=0$ Solving for V:

V = 3.3m/s and V=-1.514m/s Replacing this value into one of our previous x or y-axis equations:

$\beta =acos((V*cos\alpha )/(3) ) = 10.29°$

The amount of time:

$t = D/V = (1118m)/(3.3m/s) =338.7s$

A magnet of mass 0.20 kg is dropped from rest and falls vertically through a 35.0 cm copper tube. Eddy currents are induced, causing the copper to warm up. The speed of the magnet as it emerges from the tube is 1.10 m/s. How much heat energy is dissipated to the environment?

Answers

Answer:

0.58 J

Explanation:

We know that Total energy is conserved.

Initial Kinetic energy + Initial potential energy = final kinetic energy+ final potential energy + dissipated heat energy

Initial kinetic energy = 0 ( magnet is at rest initially)

Initial Potential energy = m g h = (0.20 kg)(9.81 m/s²)(0.35 m) = 0.69 J

Final kinetic energy = 0.5 m v² = 0.5 ×0.20 kg × 1.10 m/s = 0.11 J

Final potential energy = 0

∴ Dissipated heat energy = (0.69 -0.11) J = 0.58 J

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.

Answers

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_(i)$ = 60 mph = 26.8224 m/s

Final velocity $V_(f)$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $(1)/(2)$ m( $V_(i)$ ² - $V_(f)$ ² )

we substitute

Δk = $(1)/(2)$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $(1)/(2)$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

The switch in the circuit has been in the left position for a long time. At t=0 it moves to the right position and stays there.a. Write the expression for the capacitor voltage v(t), fort≥0. b. Write the expression for the current through the 2.4kΩ resistor, i(t), fort≥0+.

Answers

Answer:

Pls refer to attached file

Explanation:

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