Donny, who has a mass of 75.0 kg, is riding at 25.0 m/s in his truck when he suddenly slams on the brakes to avoid hitting a squirrel crossing the road. His seatbelt brings his body to a stop in 0.500 seconds. What force does the seatbelt exert on him?

Answers

Answer 1

Answer:

F = 3750  N

Explanation:

Given that,

Mass of Donny, m = 75 kg

Initial speed, u = 25 m/s

He suddenly slams on the brakes to avoid hitting a squirrel crossing the road, final speed = 0

Time, t = 0.5 s

We need to find the force the seatbelt exert on him. The force is given by :

F = ma

a is acceleration

[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{75\times (0-25)}{0.5}\\\\=-3750\ N[/tex]

So, the force is 3750  N.

Answer 2

The force the seatbelt exert on him is 3750 N

Using Newton's third law

f = ma

where

m = mass

a = acceleration

a = v - u / t

where

v = final velocity

u = initial velocity

t = time

Therefore,

f = m(v - u / t)

m = 75 kg

u = 25 m/s

t = 0.500 secs

v = 0 m/s (The car will stop because he applied a brake)

f = 75 × (0 - 25 / 0.5)

f = 75 × -25 / 0.5

f = 75 × -50

f = - 3750

f = 3750 N

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Related Questions

Convert 45km to miles show your work

Answers

Answer:

distance = 27.95 [mi]

Explanation:

in order to solve this problem, we must use the appropriate conversion factor, i.e. a conversion factor that relates the kilometers to Miles.

[tex]1 [km] = 0.6214 [mill]\\45[km]*0.6212[\frac{mill}{1km} ]=27.95 [mill][/tex]

6th grade science I mark as brainliest.​

Answers

Answer:

Solution:-

Distance =400m

Time=20s

We need to find speed

As we know that

[tex]{\boxed{\sf speed \dfrac {Distance {}_{(d)}}{Time {}_{(t)}}}}[/tex]

Substitute the values

[tex]\LARGE\leadsto\sf speed=\dfrac {400}{20}[/tex]

[tex]\LARGE\leadsto\sf 20m/s[/tex]

What two things does force depend on​

Answers

Hey Um think it’s Mass and distance

A 52 kg person is running 2.5 m/s. What is their momentum?

Answers

Answer:

130kg/ms

Explanation:

given data mass,52kgvelocity,2.5m/smomentum?from momentumM=mv

Mass=52kg

Velocity=2.5m/s

Momentum=?

We know that,

Momentum=Mass × velocity

Or,momentum=52kg× 2.5m/s

Or,momentum=130 kgm/s

So,the momentum is 130 kgm/s.

Two students are studying for a Science Test. On the table, they had a glass of water and identical cell phones. Both of their phones had run out of battery, therefore they were both charging in the same wall outlet. One student’s charging cord was 3ft and the other student’s charging cord was 9ft. which of the phones charged the fastest using Ohm’s Law. Cite evidence to support your theory.

Answers

Answer:

The student with the 3ft charger

Explanation:

Because the electricity from the 3ft charger has a shorter distance to mooves than the 9ft one

What is the period of a rotating object if it spins 24 times in 13 seconds?

Answers

Answer:

The period is 0.54 seconds

Explanation:

Period (T)

Is the time required for a rotating object to make one complete revolution around a circular path.

If it takes a time t to complete n revolutions, the period is:

[tex]\displaystyle T=\frac{t}{n}[/tex]

The rotating object spins n=24 times in t=13 seconds, thus its period is:

[tex]\displaystyle T=\frac{13}{24}[/tex]

T = 0.54 sec

The period is 0.54 seconds

according to newtons third law of motion what happened to the back of a skateboard when the person pushes down on the front of the skateboard

Answers

Answer:I think it would start moving

Explanation:

Which of the following are possible non-SI units to describe the rate of
acceleration? (Choose all that apply)

1) meters per second per minute

2) miles per kilometer per
second

3) miles per hour per second

4) kilometers per hour squared

5) meters per hour per kilometer

6) kilometers per minute

Answers

The possible non-SI units to describe the rate of acceleration is the kilometers per hour squared. That is option 4.

What is acceleration?

Acceleration is the relationship that exists between velocity and time. It is defined as the change of velocity in both speed and direction with respect to time.

The formula for acceleration is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt.

Velocity is defined as the rate of objects position with time.

The SI units of velocity = meter/seconds

Substitute for velocity in first equation:

a = meter/sec/sec

a = meter/sec²

But 1000meters = 1 Km

3600 secs = 1 hour

Therefore, the possible non-SI units to describe the rate of acceleration is the kilometers per hour squared.

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tarzans mass is 75kg. calculate his weight

Answers

Answer:

75kg=165.346697lbs

Explanation: F A T

The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the asphalt of the road is 0.75. What is the maximum static friction force?

Answers

Answer:

11250 N

Explanation:

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

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

Friction and normal force are related by the following equation:

F = μR

Where:

F is the frictional force.

μ is the coefficient of static friction.

R is the normal force.

With the above formula, we can calculate the frictional force acting on the car as follow:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

F = μR

F = 0.75 × 15000

F = 11250 N

Therefore, the frictional force acting on the car is 11250 N

When you mix a solution, the mass of the substances before and after you mix them should be the same.true or false

Answers

Answer:

True

Explanation:

The mixing substances can be taken to be the reactants and the resulting mixture as the product. When the substances change state during mixing, the mass of the substances does not change meaning that the mass of the product will equal the mass of the reactants .This is true due to the law of conservation of mass.

Answer:

TRUE

Explanation:

because blablablabla blablolbleep scientific talk

At the local destruction derby a 400 kg Toyota moving at 10 m/s collides with an 800 kg Chevy. Both are at rest after the collision. What was the velocity of the Chevy before the collision?​

Answers

Answer:

[tex]u_2 = -5m/s[/tex]

Explanation:

Given

Before Collision

Toyota

[tex]mass = m_1 = 400kg[/tex]

[tex]iniital\ velocity = u_1 =10m/s[/tex]

Chevy

[tex]mass = m_2=800kg[/tex]

[tex]initial\ velocity = u_2 = ??[/tex]

After Collision

Both Toyota and Chevy

[tex]final\ velocity = v = 0m/s[/tex]

Required

Determine the initial velocity of Chevy

This question will be answered using the following law of conservation of momentum which states that:

[tex]m_1u_1 + m_2u_2 = (m_1 + m_2)v[/tex]

Substitute values for m1, m2, u1 and v

[tex]400 * 10 + 800 * u_2 = (400 + 800) * 0[/tex]

[tex]4000 + 800u_2 = (1200) * 0[/tex]

[tex]4000 + 800u_2 = 0[/tex]

Collect Like Terms

[tex]800u_2 = 0 - 4000[/tex]

[tex]800u_2 = -4000[/tex]

Divide through by 800

[tex]\frac{800u_2 = -4000}{800}[/tex]

[tex]u_2 = \frac{-4000}{800}[/tex]

[tex]u_2 = -5m/s[/tex]

The velocity of Chevy before collision was 5m/s in the opposite direction of Toyota

a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.

Answers

Answer:

Average speed = ( 2V + V1 + V2)/4

Explanation:

Given that a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.

Since the distance is covered at equal intervals of time, and

Speed = distance/time

For the first half distance,

V = distance/t

Cross multiply

Distance = Vt

For the second half distance

(V1 + V2)/2 = distance/t

Distance = t(V1 + V2)/2

The average speed = total distance/ total time.

Average speed = [Vt + t( V1 + V2)/2] ÷ 2t

Average speed = (2Vt + V1t + V2t)/4t

Average speed = t( 2V + V1 + V2)/4t

Time t will cancel out

Average speed = ( 2V + V1 + V2)/4

PLEASE HELP ME IM TIMED

Answers

Yeah it’s the mantle

Find the kinetic energy of a 3.5 kg cat that has a momentum of 0.22 kg • m/s.
A) 3.98 x 10^-5 J
B) 4.32 x 10^-4 J
C) 6.91 x 10^-3 J
D) 7.22 x 10^-2 J

Answers

Answer:

6.19*10^-3 J

Explanation:

Just took the test. Kiss me

How much heat in kcal is required to change 0.5 kg of ice, originally at - 10 0 * C into steam at 110 C?Constants needed in the problemLatent heat of fusion=79.7 kcal/kg Specific heat of ice=0.5 kcal/kg/K ; Latent heat of vaporization ation = 539 kcal/kg ; Specific heat of water 1.0 kcal/kg/K Specific heat of ieam=0.480 kcal/kg

Answers

Answer:

Q = 364.25 kcal

Explanation:

In this question, we will have to calculate the heat absorptions for different steps of temperature rise and phase change. And then we will ad them to calculate total heat absorbed.

1. RISE IN TEMPERATURE OF ICE:

First, the temperature of ice will be increased from - 10°C to 0 °C. Heat absorbed during this process will be given as:

Q₁ = mC₁ΔT₁

where,

Q₁ = Heat absorbed while increasing temperature of ice = ?

m = mass of ice = 0.5 kg

C₁ = specific heat of ice = 0.5 kcal/kg k

ΔT₁ = change in temperature of ice = 0 - (-10) = 10 k

Therefore,

Q₁ = (0.5 kg)(0.5 kcal/kg.k)(10)

Q₁ = 2.5 kcal

2. MELTING OF ICE:

Now, the melting of ice will occur at 0°C and the heat absorbed during this process will be:

Q₂ = m(Latent Heat of Fusion of Ice)

where,

Q₂ = heat Absorbed during melting of ice = ?

Therefore,

Q₂ = (0.5 kg)(79.7 kcal/kg)

Q₂ = 39.85 kcal

3. RISE IN TEMPERATURE OF WATER:

Now, the temperature of water will be increased from 0°C to 100 °C. Heat absorbed during this process will be given as:

Q₃ = mC₃ΔT₃

where,

Q₃ = Heat absorbed while increasing temperature of water = ?

m = mass of water = 0.5 kg

C₃ = specific heat of water = 1 kcal/kg k

ΔT₃ = change in temperature of ice = 100 - 0 = 100 k

Therefore,

Q₃ = (0.5 kg)(1 kcal/kg.k)(100 k)

Q₃ = 50 kcal

4. VAPORIZATION OF WATER:

Now, the vaporization of water will occur at 100°C and the heat absorbed during this process will be:

Q₄ = m(Latent Heat of Vaporization of Water)

where,

Q₄ = heat Absorbed during vaporization of water = ?

Therefore,

Q₄ = (0.5 kg)(539 kcal/kg)

Q₄ = 269.5 kcal

5. RISE IN TEMPERATURE OF STEAM:

Now, the temperature of steam will be increased from 100°C to 110 °C. Heat absorbed during this process will be given as:

Q₅ = mC₅ΔT₅

where,

Q₅ = Heat absorbed while increasing temperature of steam = ?

m = mass of steam = 0.5 kg

C₅ = specific heat of steam = 0.48 kcal/kg k

ΔT₅ = change in temperature of ice = 110 - 100 = 10 k

Therefore,

Q₅ = (0.5 kg)(0.48 kcal/kg.k)(10 k)

Q₅ = 2.4 kcal

Hence, the total heat absorbed to change 0.5 kg of ice at - 10°C into steam at 110°C will be:

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 2.5 kcal + 39.85 kcal + 50 kcal + 269.5 kcal + 2.4 kcal

Q = 364.25 kcal

A book that weighs 5 N sits on a table. What force does the table apply to the book?

Answers

Answer:

E =F.d =[1/2]mv^2

mad = [1/2]mv^2

d= v^2/2a ,v=u+at , v^2 = [at]^2 since u=0

So d = at^2/2

F = ma= 20a= 50 , a=5/2 and t=2

so d = [5/2][2^2]/2=5

Explanation:

Every action has an equal and opposite reaction. It is an action-reaction principle. Therefore the table exerts a force of 5 N on the book in order to be in stable condition.

What is Newton's third law of motion?

Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.

Therefore the table exerts a force of 5 N on the book in order to be in stable condition.

The given data in the problem is ;

W is the weight of the book sits on table = 5N

N is the normal force on the book

From the equilibrium equation ;

Weight -Normal force on the book  =0

Weight =Normal force on the book

The normal force on the book =5N

Hence the table exerts a force of 5 N on the book in order to be in stable condition.

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*NEED THIS ANSWERED ASAP*

run this moving man simulation to get a visual understanding of all that you learned in this lesson about position velocity and acceleration during this activity you will explain these Concepts first click the introduction tab, and answer the question. set a small constant V on the velocity slider and check the velocity Vector box. click play, watch, and describe the motion in terms of what you see happening with the man's displacement, velocity and acceleration.​

Answers

Answer:

When the velocity is set to a constant value, the displacement increases with time but the acceleration remains zero

Explanation:

plato

When the velocity is set to a constant value, the displacement increases with time but the acceleration remains zero.

What is Velocity?Velocity is the directional speed of a object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.What is Acceleration?Acceleration, rate at which velocity changes with time, in terms of both speed and direction. A point or an object moving in a straight line is accelerated if it speeds up or slows down.

When the velocity is set to a constant value, the displacement increases with time but the acceleration remains zero.

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what is the mass and volume of 1000kg/m3 of water?​

Answers

Answer: The mass would be 1000m3 and the volume would be 1000kg

Explanation:

An object that is experiencing two vertical forces (upwards and a downwards) is moving downward with a constant speed. What can be concluded about the strength of the two forces?

Answers

Answer:

Both upward and downward forces are equal

Explanation:

When an object is acted upon by two vertical forces ; upward and downward, and the object moved at a constant speed, then the net force acting upon the object is zero(upward force + downward force = 0). This is because an object will only move at a constant velocity if the opposing forces cancels each other out. Since the falling object is already in motion, and maintaining a constant velocity, there is no for required to keep the body in motion and as such the Velocity at which the body is moving remains constant.

A car’s brakes decelerate it at a rate of -2.20 m/s2. If the car is originally travelling at 17 m/s and comes to a stop, then how far, in meters, will the car travel during that time?

Answers

Answer:

x = 65.68 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]v_{f}^{2}=v_{o}^{2}-2*a*x[/tex]

where:

Vf = final velocity = 0 (comes to stop)

Vo = initial velocity = 17 [m/s]

a = desaceleration = -2.2 [m/s²]

x = displacement [m]

Now replacing, we have:

[tex](0)^{2}=(17)^{2}-2*(2.2)*x\\4.4*x = 289\\x = 65.68 [m][/tex]

A :-) for this question , we should apply
2as = v^2 - u ^2
Given - u = 17 m/s
v = 0 m/s
a = -2.20 m/s^2
Solution -
2as = v^2 - u^2
2 x -2.20 x s = 0^2 - 17^2
s = 0^2 - 17^2 by 2 x -2.20
s = -( 17 x 17 ) by 2 x -2.20
( Cut minus ( - ) and ( - ) )
s = 17 x 17 by 2 x 2.20
s = 289 by 4.4
s = 65.6

.:. The car travelled 65.6 meters.

PLEASE HELP ME! IM TIMED

Answers

Answer:

Toothpaste

Because it is viscous

What is the speed of an eagle that travels 200 meters in 4 seconds? A. 800 m/s2 B. 800 m/s C. 50 m/s2 D. 50 m/s​

Answers

Answer:

The answer is D.

Explanation:

You have to apply distance formula :

[tex]distance = speed \times time[/tex]

[tex]let \: d = 200,t = 4[/tex]

[tex]200 = s \times 4[/tex]

[tex]4s = 200[/tex]

[tex]s = 200 \div 4[/tex]

[tex]s = 50 \: m {s}^{ - 1} [/tex]

Given:-

An eagle travels 200 meters in 4 seconds

To find:-

Speed of the eagle

Solution:-Here

Distance=d=200m

Time=t=4s

As we know

[tex]{\boxed{\sf Speed ={\dfrac{Distance{}_{(d)}}{Time{}_{(t)}}}}}[/tex]

Substitute the values

[tex]\qquad\quad {:}\longmapsto\sf Speed=\dfrac {200}{4}[/tex]

[tex]\qquad\quad {:}\longmapsto\sf Speed=50m/s [/tex]

[tex]\therefore{\underline{\boxed{\bf Speed\:of\:eagle=50m/s}}}[/tex]

Hence Correct option is D

Please help! Will mark brainliest!

Answers

Answer:

im pretty sure its d

Explanation:


An object starts at 16 m/s with an acceleration of 4.5 m/s? How far does it go in 9.0 seconds?

Answers

648 because 16 times 4.5 is 72 and that times 9 is 648

a motorbike can travel 1000 meters in 10 minutes calculate how car it can travel in 1 sec. ​

Answers

Answer:

1.67meter

Explanation:

if it can travel 1000 meters in 10 minutes, 10 minutes are 600 secs (10×60), 1000÷600 is 1.67

Answer:

HI

Explanation:

Charges are of two kinds—negative charges and ________________ charges. __________ amounts of both kinds of charges are found in every piece of matter.

Answers

Answer: the first is positive and second is equal

Explanation:

Answer: 1. Positive

Explanation:

Because there is negative charges and positive charges

Select three effective methods for helping a friend who is showing self harming behavior

Answers

Answer:

Avoid judging or criticizing your friend.

Let your friend know that you care and want to help

Share the ways you cope with stress and other problems in your life

I had this question

what is the KE of a 1.00 kg hammer swinging at 20.0 m/s? 200 Joules
I know the answer I just need help understanding it.

Answers

Answer:

Explanation:

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

Mass (m) = 1250 kg

Velocity (v) = 20.0 m/s.

Kinetic energy (K.E) =?

Kinetic energy is simply defined as energy possed by a body in motion. Mathematically, it is expressed as:

K.E = ½mv²

Where:

K.E is the kinetic energy

m is the mass of the object

v is the velocity of the object.

Thus, we can obtain the kinetic energy of the automobile by using the above formula as illustrated below:

Mass (m) = 1250 kg

Velocity (v) = 20.0 m/s.

Kinetic energy (K.E) =?

K.E = ½mv²

K.E = ½ × 1250 × 20²

K.E = 625 × 400

K.E = 250000 J

Therefore, the kinetic energy of the automobile is 250000 J

a cannon launches a 3.0 kg pumpkin with 110J of kinetic energy. what is the pumpkin’s speed?

Answers

Answer:

v = 8.56 [m/s]

Explanation:

The kinetic energy can be calculated by means of the following equation.

[tex]E_{k}=\frac{1}{2}*m*v^{2}[/tex]

where:

m = mass = 3 [kg]

v = velocity [m/s]

Ek = kinetic energy [J]

[tex]110 = \frac{1}{2} *3*(v^{2} )\\v^{2} = 220/3\\v=\sqrt{73.333}\\v=8.56[m/s][/tex]

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