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2 years ago

1. A buoyant force of 790 N lifts a 214 kg sinking boat.What is the boat's net acceleration?

Physics

1 answer:

enot [183]2 years ago

7 0

Answer:

The net acceleration of the boat is approximately 6.12 m/s² downwards

Explanation:

The buoyant or lifting force applied to the boat = 790 N

The mass of the boat lifted by the buoyant force = 214 kg

The force applied to a body is defined as the product of the mass and the acceleration of the body. Therefore, the buoyant force, F, acting on the boat can be presented as follows;

Fₐ = F - W

The weight of the boat = 214 × 9.81 = 2099.34 N

Therefore;

Fₐ = 790 - 2099.34 = -1309.34 N

Fₐ = Mass of the boat × The acceleration of the boat

Given that the buoyant force, Fₐ, is the net force acting on the boat, we have;

F = Mass of the boat × The net acceleration of the boat

F = -1309.34 N = 214 kg × The net acceleration of the boat

∴ The net acceleration of the boat = -1309.34 N/(214 kg) ≈ -6.12 m/s²

The net acceleration of the boat ≈ 6.12 m/s² downwards

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Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

4 0

3 years ago

The density of air is 1.3 kg/m^3. What mass of air is contained in a room measuring 2.5m x 4m x 10m? Give your answer in kg. Bra

krek1111 [17]

Mass= density x volume
1.3 kg/m^3 x ( 2.5x4x10) m^3
= 130 kg

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3 years ago

The current produced in homes is ______ current.

Lelu [443]

If any current is PRODUCED in an individual home, like with
a windmill or solar panels, it is direct current (DC).

The current SUPPLIED to homes by the local utility company
is alternating current (AC).

8 0

2 years ago

Read 2 more answers

You drag a crate across a floor with a rope. The force applied is 750 N and the angle of the rope is 25.0° above the horizontalH

frez [133]

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

$W=F\cos\theta\cdot d$

where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

$\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}$

Therefore the work done is 17673 J.

The power is given by:

$P=\frac{W}{t}$

the problem states that the time it takes is 6 s. Then:

$\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}$

Therefore the power is 2945.5 W

5 0

10 months ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago