A force used to lift a 25kg mass to a height of 12m in 5 sec does 2100J of work. How much force is used

A fast moving vehicle travelling at a speed of 25.4 m/s comes up behind another vehicle which is

ale4655 [162]

Answer:

a = 6.1 m / s²

Explanation:

For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle

Let's find the relative initial velocity of the two vehicles

v₀ = v₀₂ - v₀₁

v₀ = 25.4 - 13.6

v₀ = 11.8 m / s

the fastest vehicle

x = v₀ t + ½ a t²

The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is

x = 11.4 m

let's use the expression

v² = v₀² - 2 a x

how the vehicle stops v = 0

a = v₀² / 2x

a = $\frac{11.8^2}{2 \ 11.4}$

a = 6.1 m / s²

this velocity is directed to the left

8 0

3 years ago

Each kidney contains nephrons that filter the blood and remove waste. Please select the best answer from the choices provided. T

bekas [8.4K]

Answer: The correct answer is "True".

Explanation:

Nephron is a functional unit of kidney. Each kidney contains a millions of nephrons. It contains small blood vessels.

It filters the blood to remove excess water and the waste product from the blood. Then, it returns the cleaned blood back to rest of the body. It selectively reabsorbs the substance such as glucose. It excretes nitrogenous waste such as urea. It removes excess salt in the form of urine.

Therefore, the given statement is true.

8 0

3 years ago

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Identify the object or objects in motion in each example of kinetic energy.

Leto [7]

Answer:

A speedy train

Explanation:

5 0

3 years ago

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An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu

shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is $-1.74 rad/s^2$

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

$\omega_i = 2000 rpm$

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

$1 rev = 2\pi rad$

$1 min = 60 s$

We find:

$\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s$

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

$\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s$

Now we can find the angular acceleration, given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_i = 209.3 rad/s$ is the initial angular speed

$\omega_f = 104.7 rad/s$ is the final angular speed

t = 60 s is the time interval

Substituting,

$\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2$

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

$\omega' = \omega_i + \alpha t$

where we have

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 40 s, we find

$\omega' = 209.3 + (-1.74)(40)=139.7 rad/s$

d)

The angular displacement of the wheel in a certain time interval t is given by

$\theta=\omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 60 s, we find:

$\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad$

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

$\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

I will pick the branliest if answered these questions.

Bond [772]

1.

n₁ = index of refraction of diamond = 2.4

n₂ = index of refraction of water = 1.33

θ₁ = angle of incidence = 24 deg

θ₂ = angle of refraction = ?

using snell's law

n₁ Sinθ₁ = n₂ Sinθ₂

inserting the values

(2.4) Sin24 = (1.33) Sinθ₂

θ₂ = 47.22 deg

2.

c = specific heat = 4.18

m = mass of water = 3.5 kg

ΔT = change in temperature = 55 - 25 = 30 C

Q = heat taken

heat taken is given as

Q = m c ΔT

inserting the values

Q = (3.5) (4.18) (30)

Q = 439 J

3.

n = number of moles = 1

m = molar mass = 0.0399 kg

T = temperature = 27 K

v = average velocity

average velocity is given as

v = sqrt(3RT/m)

v = sqrt(3 x 8.314 x 275/0.0399)

v = 414.6m/s

4.

n = number of moles = 2

T = temperature = 35 C = 35 + 273 K = 308 K

U = internal energy

internal energy is given as

U = 2.5 n RT

U = 2.5 (2) (8.314) (308

U = 1.28 × 10⁴ J

5.

F₁ = 3300 N

F₂ = ?

A₁ = 0.060 m²

A₂ = 0.18 m²

Using pascal's law

F₁/A₁ = F₂/A₂

3300/0.060 = F₂/0.18

F₂ = 9900 N

6.

R = resistance of each resistance in series = 10 ohm

R' = equivalent resistance in series

equivalent resistance is given as

R' = 3 R

R' = 3 x 10

R' = 30 ohm

7.

R = resistance of each resistance in parallel = 10 ohm

R'' = equivalent resistance in parallel

equivalent resistance is given as

R'' = R/3

R'' = 10/3

R'' = 3.3 ohm

8.

for series circuit

R' = equivalent resistance in series = 30 ohm

V = Voltage applied = 60 Volts

i' = current in each resistor in series

current in each resistor in series is given as

I' = V/R'

i' = 60/30

i' = 2 A

for parallel circuit :

i = current in each resistor = V/R = 60/10 = 6 A

6 0

4 years ago

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