2. The system shown is accelerated by applying a tension Ti to the right-most cable. Assume the system is
frictionless. Solve for the tension in the cable between the blocks, T2, in terms of T. (not a).

Answer:

pressure = force / area

then pressure = 25 / 5 = "5" N/m^2

Answer:

Explanation:

The balloon would require a time of

t = d/v = 13.5/ (23.6cos38) = 0.7259...s

to travel the horizontal distance.

the vertical position relative to the throw point at that time is

h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)

h = 7.9652...

so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.

If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time

[tex]\text{Given that,}\\\\\text{Acceleration, a = -3.2 m s}^{-2}\\\\\text{Time, t = 14 sec}\\\\\text{Velocity, v = 8 m s}^{-1}\\\\\text{Initial velocity, v}_0 = ?\\\\\\v = v_0 +at\\\\\implies v_0 = v -at = 8 -(-3.2)(14) = 8 + 44.8 =52.8~ \text{m s}^{-1}[/tex]

Answer:

Explanation:

Relative to ground level it has

PE = mgh = 8(10)(5) = 400 J

Answer:

Essentially all of it

Explanation:

The potential energy was

PE = mgh = 6.36(9.81)(2.05) = 127.90278 = 128 J

ignoring air resistance, this PE converts to KE

With no rebound final velocity is zero, so Kinetic energy lost = 128 J

The horizontal distance where the block hits the floor is 3.2 m.

The given parameters:

The final velocity of the bullet-block system after the collision is calculated by applying the principle of conservation of linear momentum;

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.015(500) + 1(0) =v(0.015 + 1)\\\\7.5 = 1.015v\\\\v = \frac{7.5}{1.015} \\\\v = 7.39 \ m/s[/tex]

The time taken for the bullet-block system to fall to the floor after collision is calculated as follows;

[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.92}{9.8} }\\\\t = 0.43 \ s[/tex]

The horizontal distance where the block hits the floor is calculated as follows;

[tex]X = v_x t\\\\X = 7.39 \times 0.43\\\\X = 3.2 \ m[/tex]

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Answer:

The amount of gravitational potential energy an object has depends on its height and mass. The heavier the object and the higher it is above the ground, the more gravitational potential energy it holds. Gravitational potential energy increases as weight and height increases.

Explanation:

Answer:

Explanation:

It depends on where your reference frame for potential energy is chosen.

If a mass is above the reference origin and mass increases, potential energy will increase so the statement is FALSE

If the mass is at the reference origin and the mass increases, the potential energy stays constant at zero so the statement is FALSE

If the mass is below the reference origin and the mass increases, then the potential energy goes from a negative value to an even lower negative value so the statement in this case is TRUE

Answer:

The types of energy is bond breaking and bond forming in chemical energy.

Explanation:

During Chemical reaction energy is required either for breaking up bonds in case of reactants and building bonds to form products.

The chemical reaction in which energy is released is called exothermic reactions, which is released due to making up the bonds.

The chemical reaction in which energy is absorbed is called endothermic reactions, in which energy is absorbed for breaking up the bonds.

Answer: The answer is hupa loopa

Explanation:

Answer:

m ω^2 R = centripetal force

G M m / R^2 = gravitational force

ω^2 = G M / R^3    for the forces to be equal

d (density) = M / V = M / (4/3 R^3) = 3/4 M / R^3

or M / R^3 = 4 d / 3

Then

ω^2 = G 4 d / 3

d = 3.7 g/cm^2 = 3700 kg/m*3    converting

ω^2 = 6.67E-11 * 3700 * 4 / 3 = 32.9E-8

ω = 5.74E-4

P = 1 / f = 2 pi / ω = 6.28 / 5.74E-4 = 1.09E4 sec

10900 / 3600 = 3.04 hrs    for the minimum rotation period

Answer:

a = Δv/Δt = (0 - 20) / (5 - 2) = -6⅔ m/s²

Answer:

High winds, hail, excessive precipitation, and wildfires are forms and effects of severe weather, as are thunderstorms, downbursts, tornadoes, waterspouts, tropical cyclones, and extratropical cyclones. Regional and seasonal severe weather phenomena include blizzards (snowstorms), ice storms, and duststorms.

Explanation:High winds-  wind speeds as low as 23 knots (43 km/h) may lead to power outages when tree branches fall and disrupt power lines. Once wind exceed 135 knots (250 km/h) within strong tropical cyclones and tornadoes, homes  completely collapse,  and significant damage is done to larger buildings. Total disruption occurs once wind exceeds 175 knots (324 km/h)

Tornado- Typically look like a narrow funnel reaching from the clouds to the ground. Their wind speed goes from 65 to 250 miles per hour.

"An extreme weather condition in which we face the high speed wind in combination with heavy snow."

As for any blizzard has the normal wind speed of about 40 mph, and the visibility range reduces to less then 500 ft.

Answer: High winds, hail, excessive precipitation, and wildfires are forms and effects of severe weather, as are thunderstorms, downbursts, tornadoes, waterspouts, tropical cyclones, and extratropical cyclones. Regional and seasonal severe weather phenomena include blizzards (snowstorms), ice storms, and duststorms.

Explanation:

Answer:

v² = u² + 2gh

8.9² = 0² + 2(9.8)h

h = 4.0 m

v = at

t = 8.9/9.8

t = 0.91 s

Answer:

W = 5W

Explanation:

1W = 1J/s

W = Watts

J = Joules

s = seconds

W = J/s

W = 300J/60s

W = 5W

Mathematics is a broad subject that anyone who constantly practices by learning from first principle and from example will grow in perfection of the subject as time goes by. Hence, I strongly disagree

During my schools days, I struggled learning mathematics, at the time it was difficult, I began practising and learning from text examples, with time I got a hang of it and my grade and confidence level in the subject increased.

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Answer:

D) It cannot be determined without more information.​

Explanation:

Velocity is a vector meaning it has both magnitude and direction.

Force and acceleration are also vectors.

Without knowing the directions of each, we cannot know if the mass has a speed (scalar value) change.

We can know that the mass will have a velocity change at the rate of 2 m/s² in the direction of the applied force.

If that force is applied in the direction of initial velocity, the velocity (and speed) will increase in magnitude in the same direction.

If that force is applied opposite of the initial velocity, the  the velocity (and speed) will decrease in magnitude in the same direction. If the acceleration lasts long enough, velocity will eventually become zero and then become negative. At the same time, speed will become zero, and then increase again as speed is the absolute value of the magnitude of velocity.

If the force is applied at any other angle, both the velocity and the speed will change in both magnitude and direction.

Answer:Colloid - well compacted together but not one

Answer:

1 the digestive system thats all

Answer:

this is the answer

Explanation:

hope it's clear

pls Mark me as the brainliest pls

Answer:

1. An object is in motion when its distance from another object is changing.

2.Speed is given direction is called velocity.

Speed can be calculated.......

Answer:

Assuming the attic is ventilated to the open air the air pressure inside the house gets lower and the fan job becomes easier. It will speed up because while the pressure differential across the fan is increased, the mass of air being moved has decreased.

This can be verified by running a household vacuum cleaner in hose mode. With the hose open and air moving through it, the motor speed will be at a certain level producing a certain noise pitch . Placing your hand over the hose end will decrease the pressure in the hose, reduce the volume of air moving and you will hear an increase in motor pitch as it speeds up under lower load conditions.

The area inside the vacuum hose is akin to the inside of a well sealed house with an attic fan running.

In the physical sciences, pressure is compressive stress at quite a point within a confined fluid or the perpendicular force per unit area.

The force exerted perpendicularly to an object's surface per unit area across which that force is dispersed is termed as pressure. In relation to the surrounding pressure, gauge pressure is the pressure.

The air pressure within the house decreases and the fan's task gets simpler if the attic is aired to the outside air. It will increase because the amount of air being moved is much less, while the pressure difference across the fan is greater.

Running a home vacuum in hose mode will demonstrate this. The motor speed will be maintained at a set level, producing a specific noise pitch, with the duct open and air flowing through it. You will notice an increase in motor pitch when it speeds up under reduced load situations by placing your finger over the hose end, which will also lower the pressure in the hose and limit the amount of air moving.

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(a) The final velocity of the two vehicles if the collision was inelastic is 1.1 m/s.

(b)  For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c) The time taken to exert the given force is 0.00625 m (s).

The given parameters;

(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s[/tex]

(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)[/tex]

Apply one-directional velocity equation:

[tex]u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)[/tex]

Substitute the value of [tex]v_2[/tex] into equation (1);

[tex]12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\[/tex]

Solve for [tex]v_2[/tex];

[tex]v_2 = 12 + v_1\\\\v_2 = 12 - 9.81\\\\v_2 = 2.19 \ m/s[/tex]

Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c)

The change in the momentum of the truck is calculated as;

[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m[/tex]

The time taken to exert the given force is calculated as follows;

[tex]Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)[/tex]

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Answer:

The total amount of energy, kinetic plus potential, remains the same.

Explanation:

Answer:

hey man, think you forgot to put the questions

Explanation:

it's called a wave frequency

im pretty sure but than again I'm in Middle School but just trust me

Answer:

wave frequency i think. hope its helps

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

The wavelengths of the constituent travelling waves is calculated as follows;

[tex]L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}[/tex]

for first mode: n = 1

[tex]\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm[/tex]

for second mode: n = 2

[tex]\lambda = \frac{2L}{2} = L = 100 \ cm[/tex]

For the third mode: n = 3

[tex]\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm[/tex]

For fourth mode: n = 4

[tex]\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm[/tex]

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

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Answer: Time (days) = 88, the (mass) = 0.2180

Explanation:

Time (days) = 88, the (mass) = 0.2180

Answer:

Cool. What’s the Question?

Explanation: