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2 years ago

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In a track and field event, a hammer thrower accelerates the hammer (mass = 7.30 kg) from rest within four full turns (revolutio

ns) and releases it at a speed of 22.0 m/s. Assuming a uniform rate of increase in angular velocity and a radius of 1.30 m, calculate the angular acceleration.

1 answer:

lubasha [3.4K]2 years ago

3 0

Answer:

5.69755 rad/s²

Explanation:

r = Radius = 1.3 m

v = Velocity of the hammer = 22 m/s

n = Number of revolutions = 4

Angular displacement

$\theta=n\times 2\pi\\\Rightarrow \theta=4\times 2\pi\\\Rightarrow \theta=8\pi$

$\omega_i$ = Initial angular speed = 0

Final angular speed

$\omega_f=\frac{v}{r}\\\Rightarrow \omega_f=\frac{22}{1.3}\\\Rightarrow \omega_f=16.92307\ rad/s$

Angular acceleration

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{16.92307^2-0^2}{2\times 8\pi}\\\Rightarrow \alpha=5.69755\ rad/s^2$

Angular acceleration is given by 5.69755 rad/s²

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Anettt [7]

Answer:

See the explanation below.

Explanation:

Solving the first image question:

C ) The resulting force is defined by Newton's second law which tells us that the sum of the forces on a body is equal to the product of mass by acceleration. That is, there must be a force that acts on a body to produce an acceleration. If there is no acceleration it is because there are no external forces or developed by the body. And if there is no acceleration the body moves at a constant speed, in a straight line, so the response is C.

For the second image, we must remember that weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

W = weight [N]

m = mass [kg]

g = gravity acceleration [m/s²]

Now we have

m = 50 [kg]

ge = Earth gravity acceleration = 10 [m/s²]

gp = Distant planet gravity acceleration = 4 [m/s²]

We = ge*m

We = 10*50 = 500 [N]

Wp =gp*m

Wp = 4*50 = 200 [N]

Therefore the answer is D

For the third image, The mass is always going to be preserved, regardless of where the body or object is in space, its weight is the only one that changes since the gravitational force is modified. That is, the mass on the moon and on Earth will always be the same.

m = 70 [kg]

First, we must calculate the acceleration, by means of the following equation of kinematics.

$v_{f} =v_{o} +a*t$

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 0 (because stars from the rest)

a = acceleration [m/s²]

t = time = 4 [s]

20 = 0 + a*4

20 = 4*a

a = 5 [m/s²]

Now using Newton's second law which tells us that the total force acting on a body is equal to the product of mass by acceleration.

F = m*a

where:

F = force [N] (units of Newtons)

m = mass = 2 [kg]

a = acceleration = 5 [m/s²]

F = 2*5

F = 10 [N]

The body of Figure D, since a total force of 25 [N] to the left acts on it, in the rest of cases the force is zero or much less than 25 [N]

50 + 40 - 35 - 30 = F

F = 25 [N]

8 0

2 years ago

A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what woul

topjm [15]

Answer:

increase in temperature of water is 10° C

Explanation:

Given data

pizza = 500 kcal = 500000 calories

cold water = 50 L

to find out

increase in temperature of water

solution

we know heat formula that is

heat = m × specific heat × Δt

here m is mass = 50000 gram and Δt is change in temperature

and specific heat = 1 cal / gram C

so put here all value and find Δt

500000 = 50000 × 1 × Δt

Δt = 10° C

so increase in temperature of water is 10° C

4 0

2 years ago

Read 2 more answers

Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the

mash [69]

Answer:

The value is $} N = 66 \ blocks$

Explanation:

From the question we are told that

The weight of the block is $W_b = 100 \ N$

The dimension of the block is $d = 0.400 m \ X \ 0.250 \ m \ X \ 0.130 \ m$

Generally two atmosphere is equivalent to

$P_{2atm} = 2 * 1.013 *10^{5} = 202600 \ N/m^2$

Generally 1 atm = $1.013 *10^{5} N/m^2$

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

So

$A = 0.250 * 0.130$

=> $A = 0.0325 \ m^2$

Generally the force due to this blocks is mathematically represented as

$F = N * W_b$

Here N is the number of blocks

So

$} 202600 = \frac{N * 100 }{ 0.0325}$

=> $} N = 66 \ blocks$

3 0

3 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

2 years ago

An airplane weighing 11,000 N climbs to a

Gennadij [26K]

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

$W=(mg)\Delta h$

where

mg = 11,000 N is the weight of the airplane

$\Delta h = 1.6 km = 1600 m$ is the change in height

Substituting,

$W=(11,000)(1600)=17.6\cdot 10^6 J$

Now we can calculate the power delivered, which is given by

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

where

$W=17.6\cdot 10^6 J$ is the work done

$t=9.8 min \cdot 60 = 588 s$ is the time taken

Substituting,

$P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W$

Finally, we can convert the power into horsepower (hp), keeping in mind that

$1 hp = 746 W$

Therefore,

$P=\frac{2.99\cdot 10^4}{746}=40.1 hp$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0

3 years ago