a student lifts a 15N mass through a distance of 1.5m. whats the works done ?

Answer:

B. 25 seconds

Explanation:

500÷20=25

Answer:

7.01yard/sec

Explanation:

Given parameters:

Initial position  = 50yard

Final position = 12yard

Time  = 5.42s

Unknown:

Average speed of runner  = ?

Solution:

To solve this problem;

        Speed  = [tex]\frac{distance}{time}[/tex]  

Distance covered  = Initial position  - final position  = 50 - 12  = 38yards

So;

       Speed  = [tex]\frac{38}{5.42}[/tex]   = 7.01yard/sec

Answer:

[tex]10\: \mathrm{J}[/tex]

Explanation:

The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.

The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].

After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.

To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]

We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].

To find acceleration:

[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].

Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].

Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].

Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).

The change in the kinetic energy of the car is 10 J.

The given parameters;

The acceleration of the car is calculated as follows;

[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]

The final velocity of the car is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]

The change in the kinetic energy of the car is calculated as follows;

[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]

Learn more about kinetic energy here: https://brainly.com/question/1932411

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:

0.01

Explanation:

Given the data:

10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90

True value = 9.81

Mean value :

Σx / n

Sample size, n = 9

(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9

= 88.69 / 9

= 9.854

Standard deviation (σ) :

Sqrt (Σ(X - m)² / n)

[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9

Sqrt(0.113824 / 9)

Sqrt(0.0126471)

σ = 0.1124593

Standard Error = σ / sqrt(n)

Standard Error = 0.1124593 / 9

Standard Error = 0.0124954

Standard Error = 0.01 ( 1 significant digit)

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.  

To learn more about soundwave, here:

https://brainly.com/question/31851162

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Answer:

See explanation

Explanation:

Now we have;

vi = (-30, 0) m/s

vf = (0, 20) m/s

Δvav= vf - vi= (30, 20) m/s

magnitude of Δvav= √30^2 + 20^2 = 36.0 m/s

Direction = tan-1(20/30) = 33.69°

For aav

aav= Δvav/t = (30/5, 20/5) = (6,4) m/s^2

magnitude of  aav = √6^2 + 4^2 = 7.2 m/s^2

direction of aav = tan-1(4/6) = 33.69°

Answer:

S = 16.3125m

Explanation:

Given the following data;

Acceleration, a = 14.5m/s²

Time, t = 2.25secs

Since the bottle rocket starts from rest, its initial velocity is 0m/s.

To find the distance S, we would use the second equation of motion.

S = ut + ½at²

Substituting into the equation, we have

S = 0(2.25) + ½*14.5*2.25

S = 0 + 7.25*2.25

S = 16.3125m

Therefore, the bottle rocket covered a distance of 16.3125 meters.

Answer:

[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]

Explanation:

Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].

However, it's also given as change in momentum (impulse-momentum theorem).

Therefore, we can set the change in momentum equal to the former formula for impulse:

[tex]\Delta p=F\Delta t[/tex].

Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:

[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].

Now we plug in our values and solve:

[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).

Answer:

Analysis

Explanation:

Because you must Analysis each and every cold too find out which virus caused this.

It’s weird because Interpretation and Analysis have the meaning of examination

Answer:

the correct answer is B

Explanation:

Answer:

.B

Explanation:

This is the answer on edge 2021

Have a good day!

Answer:

a. F = 245 Newton.

b. Workdone = 392 Joules.

c. Power = 196 Watts

Explanation:

Given the following data;

Mass = 25kg

Distance = 1.6m

Time = 2secs

a. To find the force needed to lift the mass (in N );

Force = mass * acceleration

We know that acceleration due to gravity is equal to 9.8

F = 25*9.8

F = 245N

b. To find the work done by the student (in J);

Workdone = force * distance

Workdone = 245 * 1.6

Workdone = 392 Joules.

c. To find the power exerted by the student (in W);

Power = workdone/time

Power = 392/2

Power = 196 Watts.

Answer:

the car has greater momentum.

the car has greater kinetic energy.

Explanation:

FOR MOMENTUM:

Momentum is given as the product of mass and velocity of an object:

[tex]P = mv[/tex]

where,

P = momentum

m = mass

v = velocity

For Truck:

[tex]P_{truck} = (3530\ kg)(21\ m/s)\\P_{truck} = 74130 Ns[/tex]

For Car:

[tex]P_{car} = (1620\ kg)(54\ m/s)\\P_{car} = 87480\ Ns[/tex]

Therefore, car has greater momentum.

FOR KINETIC ENERGY:

Kinetic Energy is given as:

[tex]K.E = \frac{1}{2} mv^{2}[/tex]

where,

K.E = Kinetic Energy

m = mass

v = velocity

For Truck:

[tex]K.E_{truck} = \frac{1}{2} (3530\ kg)(21\ m/s)^{2}\\K.E_{truck} = 778365\ J = 778.36\ KJ[/tex]

For Car:

[tex]K.E_{truck} = \frac{1}{2} (1620\ kg)(54\ m/s)^{2}\\K.E_{truck} = 2361960\ J = 2361.96\ KJ[/tex]

Therefore, car has greater kinetic energy.

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:

They can cause sunburn. Exposure to UV rays can cause premature aging of the skin and signs of sun damage such as wrinkles, leathery skin, liver spots, actinic keratosis, and solar elastosis. UV rays can also cause eye problems.

Explanation:

Answer:

every number to 3 sf = 1) 45.0  2) 250  3) 1.30

Explanation:

your welcome :)

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:

See explanation

Explanation:

Centripetal force is defined as the inward force required to keep an object moving with a constant speed in a circular path.

The magnitude of this force depends on the mass of the object, radius of the object and the velocity of the body.

So we can write;

F = mv^2/r

Answer:

v= -107.8 m/s

Explanation:

       [tex]v_{f} = v_{o} + a*t = a*t = -g*t = 9.8m/s2*11s = -107.8 m/s (1)[/tex]

Answer:

it is defined as splitting up the given force into a number of components, without changing its effects on the body is called resolution of forces. A force is generally resolved along with two mutually perpendicular directions.

Explanation:

if we hold a very hot glass tumbler under cold water, it cracks. This is because the outer surface of the glass comes in direct contact with cold water and contracts more as compared to the inner surface. We observed that water expanded on heating.

Railway tracks consist of two parallel metal rails joined together. Small gaps, called expansion gaps, are deliberately left between the rails as there is an expansion of the rails in hot weather. Water expands on heating.

Answer:

1. 14 g of chocolate mixture.

2. 24 fl oz of chocolate milk

3. 10 cups of chocolate milk.

4. 12½ cups.

Explanation:

From the question given above, the following data were obtained:

1 TBSP = 7 g

1 Cup = 8 fl oz

2 Table spoons (TBSP) for 1 cup (8 fl oz) of milk.

1. Determination of the mass of chocolate mixture in 1 cup of chocolate milk.

From the question given above,

1 Cup required 2 Table spoons (TBSP)

But

1 TBSP = 7 g

Therefore,

2 TBSP = 2 × 7 = 14 g

Thus, 1 Cup required 14 g of chocolate mixture.

2. Determination of the number fl oz of chocolate milk in 3 cups

1 Cup = 8 fl oz

Therefore,

3 Cups = 3 × 8

3 Cups = 24 fl oz

Thus, 24 fl oz of chocolate milk are in 3 cups.

3. Determination of the number of cups of chocolate milk produce from 20 TBSP.

2 TBSP is required to produce 1 cup.

Therefore,

20 TBSP will produce = 20/2 = 10 Cups.

Thus, 10 cups of chocolate milk produce from 20 TBSP.

4. Determination of the number of cups obtained from 100 fl oz chocolate milk.

8 fl oz is required to produce 1 cup.

Therefore,

100 fl oz will produce = 100 / 8 = 12½ cups.

Thus, 12½ cups is obtained from 100 fl oz chocolate milk.

Answer:

the total distance is 5km and the displacement is 1km

Explanation:

The total distance would be the addition of John running both ways so 3 km, 2 km.

However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.

Think about 2 km as a positive value for the first part of the question and a negative value for the second part.