I need help with number 3!!!!!!!!!!!!​

Answer:

5m/s²

Explanation:

Given parameters:

Mass of refrigerator  = 50kg

Applied force  = 300N

Frictional force  = 50N

Unknown:

Acceleration of the refrigerator  = ?

Solution:

To solve this problem:

  Net force = m x a

  m is the mass

   a is the acceleration

 Net force  = Applied force  - Frictional force

             300  - 50  = 50 x a

                   250  = 50 x a

                    a  = 5m/s²

Answer:

Explanation:

The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the  Ω symbol.

100 mmHg

Givens

V1 = 20 cm^3

V2 = 80 cm^3

P1 = 400 mmHg

P2 = ?

Formula

V1 * P1 = V2 * P2

Solution

20 * 400 = 80 * P2              Divide by 80

20 * 400/80 = P2

P2 = 8000 / 80

P2 = 100 mmHg

Answer:

6692J

Explanation:

Power is defined as the rate at which work is being done.

So,

    Power  = [tex]\frac{workdone}{time }[/tex]  

  Work done  = Power x time

Given parameters:

Power  = 478watts

Time  = 14s

So;

 Work done  = 478 x 14  = 6692J

Answer:

A battery works because of the electrical energy inside of the battery, which is connected through a wire, which transmits energy through the metal inside of the wire to power to object that the wire is connected to.

Explanation:

Explanation:

Current output at the battery will be current of entire circuit, while the current through each bulb in the parallel circuit is the total current circuit.

So, current output through power supply is i and current through each component be [tex]i_1, i_2 , i_3[/tex] considering only three component.

Then in a parallel circuit

[tex]i = i_1+i_2+i_3[/tex]

The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.  

Closeness to the Sound Source: When a rocket is fired on the launchpad, it creates a tremendous amount of noise in proximity to the nearby equipment and structures.

When a rocket is launched, concentrated sound waves are created that can seriously harm neighboring structures, especially if such structures are not built to handle such strong vibrations. In contrast, once a rocket is in the air, the sound waves spread out and become less forceful as they travel through the atmosphere, decreasing the possibility that they may cause harm.

Reflection and Amplification: The launchpad environment can serve as an echo chamber for sound waves because of its huge, solid structures.

Hence, The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.  

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Answer:

6.4m/s

Explanation:

The total mechanical energy of the man is 1780J.

This mechanical energy is the energy due to the motion of the body and it is a form of kinetic energy.

 Also, mass  = 87kg

   Kinetic energy  = [tex]\frac{1}{2}[/tex]   m v²

m is the mass

v is the velocity

           1780  =  [tex]\frac{1}{2}[/tex] x 87 x v²

                v²  = 40.9

                v  = 6.4m/s

Answer:

B.

Explanation:

Answer:

0.00547s

Explanation:

Step one:

given data

mass= 0.4kg

force= 1060.9N

velocity = 14.5m/s

Frome

Ft= mv

substitute

t= mv/F

t=0.4*14.5/1060.9

t=5.8/1060.9

t=0.00547s

Answer:

15 seconds

Explanation:

If car was moving at 20m/s for 3 sec.

if car traveled 100m = 15 sec total

Answer:

6.79 m/s

Explanation:

By applying the principle of conservation of momentum.

The total momentum = MV - mv = 0 (since the squid is beginning at rest)

the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg

= 4.75 kg

speed of the squid (V) = 2.5 m/s

mass of the water expelled (m) = 1.75 kg

speed of the water (v) = ???

∴

4.75 × 2.5 = 1.75 × v

[tex]v = \dfrac{4.75 \times 2.5}{1.75 }[/tex]

v = 6.79 m/s

Answer:

D. wave one has a lower pitch

Explanation:  because The sound an object makes changes depending on how fast it is vibrating. When an object vibrates quickly, high-pitched sounds are heard. Low-pitched sounds come from things that vibrate more slowly.

Answer:

som,eythiung did cant wait

Explanation:

ya so i texted maria and she said i con c()me over so letyeyey yeah betty letty

PAINN I MISS JUOICE WORLDD PAINNNN

Answer:

Not enough detail as it is very defendant on the person and a bunch of factors in health, but overall your heart rate shouldn't reach 170 as an adult walking at a moderate intensity level, that would be closer to extreme intensity.

Explanation:

Answer:

B

Explanation:

on edge

The illustration with a vector pointed right going through an opening in a boundary and turning into a transverse wave on the other side demonstrates the phenomenon of absorption, so, option B is correct.

Absorption, in wave motion, is the process by which a wave's energy is transferred to matter when the wave travels through it. The energy of an electromagnetic, acoustic, or other wave is related to the square of its amplitude, which is the maximum displacement or movement of a point on the wave.

The amplitude of a wave continuously diminishes as it travels through a substance. The medium is described as being transparent to a specific type of radiation if just a tiny portion of the energy is absorbed, whereas it is described as opaque if all the energy is lost.

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It will accelerate. A force of magnitude F exerted on an object of mass m applies an acceleration a according to

F = m a

Remember that force and accleration are vector quantities, so the object's acceleration would point in the same direction as the applied force. The other choices describe some possible outcomes, but

• it the object starts in motion, it can only stop moving if the force opposes the motion and acts in the opposite direction. The object would eventually stop, but only for an instant before changing direction and starting to move again. Consider a ball being thrown directly upward, reaching its highest point, then falling again.

• its speed (which is a scalar quantity) would only decrease if the object starts in motion and is slowed down. But if it starts at rest, its speed can only increase. Consider a ball being dropped from some height and gaining speed as it falls.

• its velocity will certainly change, but can only increase if the object is at rest, or if it's already in motion and the force points in the same direction.

Answer:

It will accelerate.

Explanation:

i go to K12

Answer:

why compasses point towards the north and south poles of the earth.

Explanation:

hope it helps!!

Answer:

a)   a = 4,552 m / s²,   b)  a = 2,588 m / s²

Explanation:

Newton's second law is

          F = ma

          a = F / m

in this case the force remains constant

indicate us

* for a mass m₁

       a₁ = F/m₁

       a₁ = 12, m/ s²

* for a mass m₂

        a₂= 3.3 m / s²

a) acceleration

       m = m₂-m₁

we substitute

        a = [tex]\frac{F}{m_2 - m_1}[/tex]

        1 / a = [tex]\frac{m_2}{F} - \frac{m_1}{F}[/tex]

let's calculate

        [tex]\frac{1}{a}[/tex] = [tex]\frac{1}{3.3} - \frac{1}{12}[/tex]

         [tex]\frac{1}{a}[/tex] = 0.21969

         a = 4,552 m / s²

b)   m = m₂ + m₁

     a = F / (m₂ + m₁)

     [tex]\frac{1}{a} = \frac{m_2}{F} + \frac{m_1}{F}[/tex]

we substitute

           [tex]\frac{1}{a} = \frac{1}{3.3} + \frac{1}{12}[/tex]

           a = 2,588 m / s²

Answer:

Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.

Explanation:

eventually the crust cracks.

Answer:

1.007 kg/m³

Explanation:

Given that:

volume = 2500 m³

density of air [tex]\rho = 1.23 \ kg/m^3[/tex]

weight of air displaced = [tex]V \times \rho \times g[/tex]

= 2500 × 1.23 × 9.81

= 30165.75 N

weight of the ballon fabric = 860 N; &

The propane weight = 1200 N

The passenger additional weight = 3400 N

weight of the heated gas will be = V × d × g

= 2500 × d  9.81

= 24525 d

For floating

The weight of air displaced is less than 30165.75

∴ 860 + 1200 + 3400 + 24525 d = 30165.75

5460 + 24525 d = 30165.75

24525 d = 24705.75

d = 24705.75  / 24525

d = 1.007 kg/m³

Hence, the average density of the heated gas = 1.007 kg/m³

Answer:

the tack's tangential speed is  5.59 m/s

Explanation:

Given that;

R = 0.331 m

wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so

ω = 2.69 rev/s × 2π/1s = 16.9 rad/s

using the relation of angular speed with tangential speed

tangential speed v of the tack is expressed as;

v = R × ω

so we substitute

v = 0.331 m × 16.9 rad/s

v = 5.59 m/s

Therefore, the tack's tangential speed is  5.59 m/s

Answer:

California has one of the most variable climates of any U.S state, and often experiences very wet years followed by extremely dry ones . The state's reservoirs have insufficient capacity to balance the water supply between wet and dry years.

PLEASE MARK ME AS BRAINLIEST

Answer:

the answer is B

Explanation:

I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot

water is cooler, and D shows that both are cold

Answer:

[tex]500 \: \mathrm{kg} \cdot \mathrm{m/s}[/tex]

Explanation:

The momentum of an object is given as [tex]p=mv[/tex]. Since Amy has a mass of 50 kg and is travelling 10 m/s, her momentum is [tex]p=mv=50\cdot 10 =\fbox{$500\: \mathrm{kg\cdot m/s}$}[/tex].

Answer:

500

Explanation:

Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

A)

       [tex]q_{g} = c_{w} * m * \Delta T (1)[/tex]

       [tex]q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)[/tex]

B)

       [tex]q_{l} = c_{Al} * m_{Al} * \Delta T (3)[/tex]

⇒    [tex]-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)[/tex]

       [tex]c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)[/tex]

Answer:

a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.

b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.

Explanation:

a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:

[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (1)

[tex]\Sigma F_{y} = N -m\cdot g = 0[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]N[/tex] - Normal force from ground to the car, measured in newtons.

[tex]v[/tex] - Maximum speed of the car, measured in meters per second.

[tex]R[/tex] - Radius of the circular track, measured in meters.

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

By applying (2) in (1):

[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex] (3)

The force of static friction the road can apply in the car ([tex]f[/tex]), measured in newtons, is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])

[tex]f = \mu_{s}\cdot m \cdot g[/tex]

[tex]f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]f = 8826.3\,N[/tex]

The force of static friction the road can apply before the car starts to move is 8826.3 newtons.

b) Then, we calculate the maximum speed of the car by (3):

[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex]

[tex]\mu_{s}\cdot g = \frac{v^{2}}{R}[/tex]

[tex]v = \sqrt{\mu_{s}\cdot g\cdot R}[/tex]

If we know that [tex]\mu_{s} = 0.6[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 200\,m[/tex], then the maximum speed of the car can travel before it would start to slide is:

[tex]v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}[/tex]

[tex]v \approx 34.305\,\frac{m}{s}[/tex]

The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.