what would the mass be of an object that was moving at a velocity of 35 m/s and has a kinetic energy of 500j be?

Answers

Answer 1

Answer:

0.816 kg

Explanation:

[tex]E_k=\frac{1}{2}mv^{2}\\[/tex]

so [tex]m=\frac{2E_k}{v^2}=\frac{2\times500}{35^2}=0.816 kg[/tex]


Related Questions

What does it mean to discover
the truth in the context of
detective work? What kinds of
truth can be discovered?

Answers

Answer:

When the machine indicates a lie, what is the probability that the suspect is really lying? If the machine does not indicate a lie, what is the probability that the suspect is really telling the truth?

My solution if the probability that a test subject is chosen at random from the 200 in fact lie = 80/(80+5)=0.94

Then when 15% of people arrested lie, the probability that the suspect is really lying is 0.15*0.94=0.141

One mole of a monatomic gas is subjected to the following sequence of steps.

a. Starting at 300 K and 10 atm, the gas expands freely into a vacuum to double its volume.
b. The gas is next heated reversibly to 500 K at constant volume.
c. The gas is then reversibly expanded at constant temperature until its volume is doubled
d. The gas is finally reversibly cooled to 300 K at constant pressure

Required:
Calculate the values of q and w and the changes in U, H, and S.

Answers

Answer:

c

Explanation:

the gas is the reversible expanded and constant temperature until it's volume is doubled

Ben and Dan both weigh 600 N. They are doing pull-ups together. Each pull-up is 0.5 meters of distance. How much work do they each do for every pull-up? 300 J 600 J 1200 J 1500 J

Answers

Answer:

300 J

Explanation:

Work = (Force)*(distance) = 600 N ∗ 0.5 m = 300 J

For the purpose of calculating the electric field strength by means of Gauss’s law, determine whether approximate cylindrical symmetry holds in each of the following situations.
Part (a) We have a 3.6-m long copper rod of radius 1 cm, carrying a charge of 1.5 nC distributed uniformly along the rod’s length. We want to calculate the electric field strength at a point 4.9 cm from the rod near its center.
TRUE FALSE
Part (b) We have a 8.9-cm long copper rod of radius 1 cm, carrying a charge of 1.5 nC distributed uniformly along the rod’s length. We want to calculate the electric field strength at a point 4.9 cm from rod near its center.
TRUE FALSE
Part (c) A 1.9-m long wooden rod is glued end-to-end to a 1.9-m long plastic rod, both of radius 1 cm. The combined rod is then painted with an electrically charged paint so that it is covered with a uniform charge density, giving it total charge of 1.5 nC. We want to calculate the electric field strength at a point 4.9 cm from the rod near its center.
TRUE FALSE
Part (d) For the same charged rod as in part (c) we want to calculate the electric field strength at a point 8.9 m from each end of the rod.
TRUE FALSE

Answers

Answer:

1. True

2. False

3. True

4. False

Explanation:

a) In this case, length of the copper rod is 3.6 m which is much larger than the distance 4.9 cm to the point at which electric field is to be determined. Therefore, yes, cylindrical symmetry holds.

b) In this case, length of the copper rod is 8.9 cm which is of the same order of magnitudes the distance 4.9 cm to the point at which electric field is to be determined. Therefore, no, cylindrical symmetry does not hold.

c) In this case, length of the copper rod is 3.8 m which is much larger than the distance 4.9 cm to the point at which electric field is to be determined. Therefore, yes, cylindrical symmetry holds.

d) In this case, length of the copper rod is 3.6 m which is of the same order of magnitudes the distance 4.9 cm to the point at which electric field is to be determined. Therefore, no, cylindrical symmetry does not hold.

Identify the region of the electromagnetic spectrum where the wavelength measurement of 7.5 nm can be found.

Answers

We have that the  electromagnetic spectrum where the wavelength measurement is 7.5 nm is the X-ray

X-ray

Option D

Question Parameters:

the wavelength measurement of 7.5 nm can be found

Generally, gamma rays has its wavelength around 10pm.

The microwaves has its wavelength around 1cm.

The ultraviolet has its wavelength around the 400nm.

The range of the X-rays is between to 5nm to 10nm.

Hence,the electromagnetic spectrum where the wavelength measurement is 7.5 nm is the X-ray

X-ray

Option D

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CQ

Identify the region of the electromagnetic spectrum where the wavelength measurement of 7.5 nm can be found.

A) Gamma rays

B) Microwaves

C) Ultraviolet

D) X-rays

E) Visible light

Х
In which layer of
the soil do we
usually find
loam? to. Topsoil
b. Parent Soil c.
Bedrock d.
ground-soil

Answers

Answer:

A. topsoil the answer

Explanation:

I think its a correct answer

A 10 kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 5 N opposing the motion.
Calculate the acceleration of the object.

Answers

Newton second's law

∑F = m.a

40 - 5 = 10.a

35 = 10.a

a = 3.5 m/s²

Answer:

The acceleration of the object was 3.5 m/s².

Explanation:

Let the direction of the horizontal force be positive.

By Newton's second law, the sum of the horizontal forces is as follows:
[tex]\displaystyle \sum F_x = F_A - F_k = ma[/tex]

Where Fk is the frictional force and FA is the horizontal force.

Substitute in known values and solve for acceleration a:

[tex]\displaystyle \begin{aligned} (40\text{ N}) - (5\text{ N}) & = (10\text{ kg})a \\ \\ a &= 3.5\text{ m/s$^2$}\end{aligned}[/tex]

Hence, the acceleration of the object was 3.5 m/s².

A car moving 12.2 m/s starts to coast up a frictionless 14.0 degree hill. what is its acceleration?

Answers

Answer:

−2.373

Explanation:

Theta=14

ma=-mgsin(theta)

a= -gsin(theta)

a=  -9.81sin(14)

a=−2.373

A car moving 12.2 m/s starts to coast up a frictionless 14.0 degree hill. -2.37 m/s² is its acceleration.

What is acceleration ?

Acceleration is the rate at which an object's velocity with respect to time changes. They are vector quantities, accelerations. The direction of the net force applied on the item determines the object's acceleration.

An object has positive acceleration if it is accelerating and travelling in the right direction. Positive acceleration was demonstrated in the first example by the speeding car. The acceleration is occurring in the same direction as the car's motion, which is forward and speeding up.

The equation a = v/t denotes acceleration (a), which is the change in velocity (v) over the change in time (t). Using this, you may gauge the speed of velocity changes in meters.

θ = 14

Acceleration = -gsinθ

= - 9.8 × sin 14°

= -2.37 m/s²

Thus, A car moving 12.2 m/s starts to coast up a frictionless 14.0 degree hill. -2.37 m/s² is its acceleration.

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#SPJ2

What is the reading of the Voltmeter shown in the figure?

a. 8V

b. 10V

c. 16V

d. 20V​

Answers

Answer:

Correct option is A)

Let us assume voltmeter is ideal:-It means voltmeter has infinite resistance and current will not flow in voltmeter.

R

eq

in upper branch = R

eq

in lower branch. So current will get equally divide into both branches.

V

A

−V

B

=4×1(1)

V

A

−V

D

=16×1(2)

Subtract 2 and 1

V

B

−V

D

=12V

A voltmeter will read 12 V

What are some of the major pieces of evidence for the theory of plate tectonics? How does the theory explain these observations?

Answers

Answer:

evidence:

- fossils found in different parts of the world were the same

- weather patterns in different parts of the world

- mountain ranged with the same rock types, structures, and ages are now on opposite sides of the atlantic ocean

- coal was found in cold climates where it could not have formed

-scratches from glaciers were found in a desert

- continents fit like a puzzle

- identical rocks of the same type and age found on both sides of the atlantic ocean

The volume
of this ice
would be
considered
what type of
property?
A. a chemical property
B. a rare property
C. a physical property
D. an undetermined property

Answers

c. a physical property

A wooden supply raft is 2 m wide, 3 m long and 0.200 m deep. The raft and it’s occupants have mass of 700 kg. How deep will the sink below the water when floating?

Answers

58.3% of the raft will sink below the water when floating.

Volume of the raft

The volume of the raft is calculated as follows;

V = Lbh

V = 2 x 3 x 0.2

V = 1.2 m³

Density of the raft

The density of the raft is calculated as follows;

ρ = m/v

where;

m is the mass = 700 kgv is the volume = 1.2 m³

[tex]\rho = \frac{700}{1.2} \\\\\rho = 583.33 \ kg/m^3[/tex]

Fraction of the raft below the water

[tex]f = \frac{\rho\ _{raft}}{\rho \ _{water}} \times 100\%\\\\f = \frac{583.33}{1000} \times 100\%\\\\f = 58.3 \%[/tex]

Thus, 58.3% of the raft will sink below the water when floating.

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A cannonball is shot (from ground level) with an initial horizontal velocity of 20 m/s and an initial vertical velocity of 30 m/s. What's the maximum height reached by the cannonball

Answers

Answer:

maximum height =  46m

Explanation:

Vx = 20m/s

Vy = 30 m/s

gravity = 9.8 m/s/s

final velocity = initial velocity - gravity • time

0 = 30m/s - 9.8m/s/s • time

subtract 30m/s from both sides

-30m/s = - 9.8m/s/s • time

divide both sides by -9.8m/s/s

3.06 s = time

height =  initial velocity • time - 1/2 • gravity • time^2

height = 30m/s • 3.06s - 1/2 • 9.8m/s/s • 3.06s ^2

height = 46m

the maximum height is equal to 46m

Another elephant pushes with 3500 N on a load of trees. It then pushes these trees for 225 m. How much work did he elephant do?

Answers

Answer:

787, 500 N/m

Explanation:

Work = Force • displacement

[tex]Work = (3500 N)(225 m) = 787500\frac{N}{m}[/tex]

[tex]W = (3500 N)(225 m) = 787500\frac{N}{m}[/tex]

A wheel rolls 5 revolutions on a horizontal surface without slipping. If the center of the wheel moves 3.2 m, what is the radius of the wheel

Answers

The radius of the wheel is 0.102 m.

What is radius of a circle?

The radius of a circle is a straight line that comes from the center of a circle to any point of the circumference of the circle.

To calculate the radius of 5 revolutions of the wheel, we use the formula below.

Formula:

d = 10πr.............. Equation 1

Where:

d = Total distance moved by the wheelr = radius of the wheel.π = pie

Make r the subject of the equation

r = d/(10π)........... Equation 2

From the question,

Given:

d = 3.2 mπ = 3.14

Substitute these values into equation 2

r = 3.2/(10×3.14)r = 3.2/31.4r = 0.102 m.

Hence, the radius of the wheel is 0.102 m.

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What happens to an object’s momentum if its mass is reduced by half but its velocity remains the same?

Answers

If the mass of an object is halved and its velocity remains the same, its momentum will be halved.

What is momentum?

Momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity.

We can calculate the momentum of an object using the following expression.

p = m × v

where,

p is the momentum.m is the mass.v is the velocity.

The initial momentum of the object is:

p₁ = m₁ × v₁

If its mass is reduced by half but its velocity remains the same, the new momentum is:

p₂ = m₂ × v₁ = 0.5 m₁ × v₁ = 0.5 p₁

If the mass of an object is halved and its velocity remains the same, its momentum will be halved.

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Help Me With This Question Please.

Thank you UwU​

Answers

8Ω, 9Ω, xΩ are in series

[tex] \therefore \: R _{1} = 8 + 9 + x = (17 + x)Ω[/tex]

3Ω and 5Ω are in series

[tex] \therefore \: R _{2} = 3+ 5 = 8Ω[/tex]

Now equivalent resistance of R1 and R2 which are parallel in R3 = 6Ω

[See The Attachment]

[tex] \therefore \frac{1}{R _{3}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} \\ \\ \therefore \: \frac{1}{6} = \frac{1}{(17 + x)} + \frac{1}{8} \\ \\ = \frac{1}{17 + x} = \frac{1}{6} - \frac{1}{8} = \frac{1}{24} \\ \\ \therefore17 + x = 24 \\ x = 24 - 17 = 7 \\ x = 7Ω[/tex]

Hope This Helps!!

The value of the resistance X is 7 ohms.

Equivalent resistance in series

The equivalent resistance in series is calculated as follows;

R₁ = 8 + 9 + x = 17 + x

R₂ = 3 + 5 = 8

Equivalent resistance in the circuit

The equivalent resistance in the circuit is calculated as follows;

[tex]\frac{1}{R} = \frac{1}{R_1}+ \frac{1}{R_2} \\\\R = \frac{R_1 R_2}{R_1 + R_2} \\\\[/tex]

[tex]R(R_1 + R_2 ) = R_1 R_2[/tex]

6(17 + x + 8) = 8(17 + x)

150 + 6x = 136 + 8x

14 = 2x

x = 7

Thus, the value of the resistance X is 7 ohms.

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2. How are the speed, wavelength, and frequency of a wave related

Answers

Answer:

speed = wavelength * frequency

Explanation:

speed = wavelength * frequency

A ball is dropped from the top of a building.
When does the ball have the least potential energy?
O after it has hit the ground
O half way through the fall
O as it is released
O just before it hits the ground

Answers

Answer:

after it has hit the ground

.........

Answer: A. after it has hit the ground

how do i explain why two balls have the same velocity if they are rolling down the same slope but they started in different positions?

Answers

Answer:

When two balls are equally weighted, no matter where they are placed on a ramp, they will travel at the same speed. One will just reach the ground quicker.

Explanation:

Question 14 (1 point)
One star has a temperature of 20,000 K and another star has a temperature of 8,000
K. Compared to the cooler star, how much more energy per second (how many times
more energy) will the hotter star radiate from each square meter of its surface?

Answers

We have that the more energy per second to radiated by the hotter  is mathematically given as

Energy per second=256

Energy per second to radiated

Question Parameters:

One star has a temperature of 20,000 K

and another star has a temperature of 8,000

Generally the equation for the Energy per second   is mathematically given as

Energy per second is Proportional to T^4

Where

T=20000/8000

T=4

therefore

Energy per second=4^4

Energy per second=256

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- At time t = 0, a small ball is projected vertically upwards from a point A which is 24.5 m above the ground. The ball first comes to instantaneous rest at the point B, where AB = 19.6 m and first hits the ground at time t = T seconds. = The ball is modelled as a particle moving freely under gravity. (a) Find the value of T.​

Answers

The value of time needed to complete the upward and the downward motion is, T=5sec .T is the time at which the ball first hits the ground.

What is the time period?

The time period is found as the ratio of the displacement and the velocity. Its unit is second and denoted by t

The given data in the problem is;

The time required to move upwards is,t₁

The time required to move downwards is,t₂

The distance traveled is,s= 19.6m

The gravitational acceleration is, g=9.81 m/s

At the highest position, the velocity of the ball is zero;

The time required to move upwards is found as;

[tex]\rm S_1 = ut + \frac{1}{2} at^2 \\\\ \rm S_1 = \frac{1}{2} (-g)t_1^2 \\\\ t_1= \sqrt{\frac{2S_1}{g} } \\\\ t_1= \sqrt{\frac{2 \times 19.6 }{9.81} } \\\\ t_1= 2 \ sec[/tex]

The time required to move downward is found as;

[tex]\rm S_2 = ut + \frac{1}{2} at_2^2 \\\\ \rm S_1 = \frac{1}{2} (-g)t_2^2 \\\\ t_2= \sqrt{\frac{2S_2}{g} } \\\\ t_2= \sqrt{\frac{2 \times 44.1 }{9.81} } \\\\ t_2= 3 \ sec[/tex]

The total time period is the sum of the time required to move upwards and the time required to move downwards.

T =t₁+t₂

T=2+3

T=5 sec

Hence the value of T will be 5 sec.

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Question 3
Which of the following is an example of positive acceleration?
A car increases its speed from 5 m/s to 10 m/s.
A car is traveling at 25 m/s for 5 minutes.
A car is traveling at 25 m/s and begins to slow down as it approaches a stop sign.
A car is in park.

Answers

Answer:    1ST OPTION   A car increases its speed from 5 m/s to 10 m/s

Explanation:IN 1ST OPTION : because the vel and  force are in the same direction therefore acceleration is positive

                     IN 2ND & 3RD OPTION  :  the force and velocity in opposite direction so its velocity decreases therefore acceleration is negitive

helphelphelphelphelphelphelphelp

Answers

Answer:

a is slower

b is constant

c is faster

What is the magnitude of the electric field strength at a position that is 1.2 m from a point charge of 4.2 \times 10^{-6} \mathrm{C}4.2×10
−6C?

Answers

Answer:

lil durk a goat

Explanation:

A pump lifts water into an overhead tank at a height of 12m at a rate of 5kg/s,what is the power in the pump?



Answers

Answer:

600 watts

Explanation:

if i release one steel ball from the top of a ramp and the other ball from the 40cm mark will they have the same acceleration?

Answers

Answer:

Yes, they will also have the same acceleration. Acceleration is controlled by the amount of weight (def. amount of gravitational pull on a given object) that the ball has. For one ball to accelerate quicker than the other, it would need a propulsion element, which it does not.

Explanation:


Activity A:
Surviving a crash
Get the Gizmo ready
• Click Reset (?).
. On the DESIGN tab, check that Sedan is selected
Introduction: Modern vehicles contain features designed to keep passengers safe in a crash.
The crumple zone in the front of the car slows the car gradually and increases stopping time.
The safety cell is a rigid cage that prevents passengers from being crushed. Inside, seat belts
and airbags prevent the driver from hitting the windshield, steering wheel, or dashboard.
Question: How does a crumple zone help protect a passenger?
1
Make a hypothesis: On the DESIGN tab, look at the parameters you can control. What
settings do you think will make the safest car? Set up the Gizmo, and then fill in below.
Crumple zone length:
Crumple zone rigidity:
Safety cell rigidity:
Seat belt present?
If present, seat belt stiffness:
Air bag present?
If present, air bag rigidity:

Answers

The crumple zone help protect a passenger from injury by converting some of the kinetic energy possessed by the car into a controlled deformation, or crumpling when there is a crash.

What is impact?

The term impact refers to a large force that acts for a short time. Passengers in a vehicles sustain injuries after an accident due to the high impact of the crash.

As such, the crumple zone help protect a passenger from injury by converting some of the kinetic energy possessed by the car into a controlled deformation, or crumpling when there is a crash.

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As a boy pulls a sled carrying his sister, he exerts a force of 2.6 lbs.at an angle of 30 degrees with the ground. How much work does he do if he pulls her at a constant speed for exactly one-half mile (2640 ft)?

Answers

The amount of work done by the boy in pulling the sled is 8064.41 J

Definition of work done

Workdone is defined as the product of force and distance moved in the direction of the force. It can be expressed mathematically as:

Workdone (Wd) = force (F) × distance (d)

Wd = Fd

Considering angle projection,

Wd = FdCosθ

How to determine the workdone Force (F) = 2.6 lb = 2.6 × 4.448 = 11.5648 NDistance (d) = 2640 ft = 2640 × 0.305 = 805.2 mAngle (θ) = 30°Workdone (Wd) =?

Wd = FdCosθ

Wd = 11.5648 × 805.2 × Cos 30

Wd = 8064.41 J

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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 100-pound weight two-feet above the ground with an energy efficiency of 25%. How many repetitions can she do with the energy supplied from a single Oreo cookie? What happens to the number of repetitions that can be done if the efficiency increases?

Answers

The number of times the athlete can lift the weight with single energy supplied by the Oreo cookie is 204 times and this number of times will increase if the efficiency increases.

Energy used by the athlete

The energy used by the athlete is calculated as follows;

E = Fd

where;

F is the applied force = 100 lb weight = 444.82 Nd is the distance = 2 ft = 0.61 m

E = 444.82 x 0.61 = 271.34 J

Input energy

The Input energy of the athlete = 53 kcal = 221752 J

Number of times the athlete can lift the weight

[tex]E = \frac{0utput \ energy}{1nput \ energy} \\\\0.25 = \frac{n(271.34)}{221752} \\\\271.34n = 55483\\\\n = \frac{55483}{271.34} \\\\n = 204[/tex]

Thus, the number of times the athlete can lift the weight with single energy supplied by the Oreo cookie is 204 times and this number of times will increase if the efficiency increases.

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