The magnitude of the frictional force of the box is 65 N
From the question given above, the following data were obtained:
Mass (m) = 25 KgForce applied (Fₐ) = 135 NNet force (Fₙ) = 70 NFrictional force (Fբ) =?The frictional force acting on the box can be obtained as follow:
Fₙ = Fₐ – Fբ
70 = 135 – Fբ
Collect like terms
70 – 135 = – Fբ
–65 = –Fբ
Multiply through by –1
Fբ = 65 NThus, the frictional force of the box is 65 N
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the image in figure 16.36 combines observations made with visible light and radio telescopes. which color in the image represents the radio emission?
Answer: pinkish- white
Purple
Explanation:
Consider a car of mass m moving with an initial speed Vo on a straight, flat road. At time t=0, the driver fully applies the breaks to avoid colliding with debris in the road in front of the car. The car’s wheels lock, causing the car to slide on the roadway until the car stops, before running over the debris. The distance that the car slides is D. The coefficient of kinetic friction between the car’s tires and the roadway is a constant value of Uk.
A) Sketch a free body diagram for the car sliding to rest and write Fnet equations
A) does the value D increase or decrease with increasing initial speed?
B) does the value D increase or decrease with increasing the coefficient of friction?
C) derive an equation for the acceleration of the car in terms of Uk and physical consonants
what was stored when you bent the ruler in your left hand?
Answer:
potential energy
the name comes from the fact that the ruler has "potential" to act
Explanation:
Explanation:
The deformation of the ruler creates a force in the opposite direction, known as a restoring force.
I'm not sure if it's correct...
1. Taking accurate measurements of small values using the ‘measure many and average’ method.
Each oscillation takes a small amount of time, and your reaction time is about 0.1 seconds. This means too much of the time you measure is your reaction time. Find out the percentage error with this calculation;
a. ((reaction time)⁄(time for one oscillation))x100=% error
b. To reduce the percentage error, time for ten oscillations.
c. Take the average time by dividing your answer by ten.
d. Your reaction time is still the same, 0.1 s. Find out the new percentage error with this calculation;
Answer:
i need free answers here
Explanation:
A position versus time graph is shown:
Position vs. Time
1 15
10
10
Time (seconds) ->
Which statement accurately describes the motion of the object in the graph above?
O It is at rest for 3 seconds before moving at constant speed for 10 seconds.
It is at rest for 3 seconds before moving at constant speed for 4 seconds.
Olt is moving from 0 cm to 5 cm at constant speed of 7 cm/s.
O it is moving from 0 cm to 3 cm at a constant speed of 3 cm/s.
Answer: Answer: The object moves forward at 5 m/s, stops, and then changes velocity.
Explanation:
With the information given in the question we can graph the points (image attached).
As we can observe, in the first segment of the graph the velocity is increasing linearly (at a constant rate) and is 5 m/s, then in the second segment we can see the position of the object remains the same from second 2 to second 4, which means the object is stopped.
Finally, in the third and last segment, we can observe a change in velocity (at a negative constant rate, because is decreasing), which is decreasing until the object stops.
Explanation:
11. A tennis ball is thrown straight up in the air, leaving the person's hand with an initial velocity of 3.0 m/s, as shown to the right. What is the displacement of the ball above the person's hand?
Answer:
0.46m
Explanation:
if clarisses mass on earth is 52kg what would be her weight on venus if the acceleration due to gravity on that planet is 8.87 m/s^2
Answer:
Weight=F=MA=52*8.87=461.24
Hi, please help! •_•
One workman is measured as having a power of 528W. His weight is 800N. He can develop the same power climbing a ladder, whose rungs are 30cm apart. How many rungs can he climb in 5s?
Answer:
He can climb 11 rungs
Explanation:
Power = 528W, Force = 800N, Time= 5s Total Distance moved = 30x cm or 0.3x m, where x is the number of rungs he climbed
Power = Work done/time
Work done = Power × time
= 528 × 5
= 2640Joules
Also Work done = Force × total distance moved
Total Distance moved = Work done/force
0.3x = 2640/800
0.3x = 33
x = 3.3/0.3
x = 11
The deceleration of the dummy is less than the deceleration of the passenger compartment.
Explain why this is of benefit for the safety of a passenger.
Based on the information given, this is beneficial because the passenger is less likely to be injured.
Deceleration simply means acceleration in a reverse way. Deceleration simply means the rate at which an object slows down.
From the complete information, it should be noted that the fact that the deceleration of the dummy is less than the deceleration of the passenger compartment is beneficial as it impacts on passenger or dummy less than without the airbag or seat belt.
Therefore, this is important as the passenger is less likely to be injured.
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2 questions, 100 points
Gravitational force between objects ____________ as objects move farther apart, and it is ___________ between objects that are closer together.
increases, weaker
increases, stronger
decreases, weaker
decreases, stronger
In science, weight is measure in
Pounds
Kelvin
Newtons
Liters
Answer:
decreases, stronger
newtons
Explanation:
You roll a toy car, and it moves 10 meters in five seconds. What is the car's velocity?
The car's velocity is 2 m/s.
Hope you could get an idea from here.
Doubt clarification - use comment section.
please help answer the question in the image!!!
Answer:
C (only)
Explanation:
Often, we consider the 0 position on a graph like this to be the starting point. However, the position on this graph is indicated as positive at time = 0. The motion shown can be described in each stage as follows.
A: moving away from the starting position toward the reference position (position = 0)
B: stationary at the reference position
C: moving away from the reference position toward the starting position
D: stationary at a point just short of the starting position
E: moving away from the starting position toward, then beyond, the reference position
The basic SI unit of length is
Answer:
m
Explanation:
Metres, m, is the SI unit of length
Hope this helped you-have a good day bro cya)
The relationship between Acceleration and Mass is described as:
O a. Equal
O b. Directly Proportional
O c. Inversely Proportional
O d. Opposite
O e. None of these
Explanation:
According to Newton's second law
[tex]\\ \sf{:}\dashrightarrow Force=Mass\times Acceleration [/tex]
[tex]\\ \sf{:}\dashrightarrow Mass=\dfrac{Force}{Acceleration}[/tex]
[tex]\\ \sf{:}\dashrightarrow Mass\propto \dfrac{1}{Acceleration}[/tex]
Option C
If a planet has an eccentricity of 0.92 and a distance of the major axis is 5,000,000 km, then what is the distance between foci?(1 point)
460,000,000 km
0.00000018 km
4,600,000 km
5,000,000 km
Answer:
it’s the third option: 4,600,000 km
Explanation:
just took the test and that was correct
a 25 kg crate is pushed horizontally with a force of 70N. if the coefficient of friction is .20, calculate the acceleration of the crate.
From Newton's second law of motion, the acceleration of the crate is 0.84m/[tex]s^{2}[/tex]
Given that a 25 kg crate is pushed horizontally with a force of 70N and the coefficient of friction is 0.20
To calculate the acceleration of the crate, We will consider the net force acting on the crate. The net force will be sum of the applied force F and the frictional force [tex]F_{r}[/tex].
Let us first calculate the frictional force [tex]F_{r}[/tex]. That is,
[tex]F_{r}[/tex] =- μN
[tex]F_{r}[/tex] = μmg
[tex]F_{r}[/tex] = 0.2 x 25 x 9.8
[tex]F_{r}[/tex] = 49N
Using Newton's second law,
F = ma
F - [tex]F_{r}[/tex] = ma
70 - 49 = 25 a
21 = 25a
a = 21 / 25
a = 0.84 m/[tex]s^{2}[/tex]
Therefore, the acceleration of the crate is 0.84m/[tex]s^{2}[/tex]
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The force of gravity can also be referred to as the force of _____________.
Answer:
Gravitational.
Explanation:
In Earth, everything have a volume and mass, or downward force of gravity, which Earth's mass exerts on them.
A weightlifter lifts a 350 N set of weights over his head a vertical distance of 2.5 meters above the floor. Calculate the work the athlete does on the weights, assuming he lifts them at a constant speed.
A. 525 J
B. 750 J
C. 1005 J
D. 700 N•m
E. 875 J
3. Which is not an example of energy
(5 Points)
light
thermal
heat
chemical
Answer:
Heat, Chemical
Answer:
heat is not an energy is a process.
Explanation:
what is the purpose of the thermal energy tranfer lab
Answer:thermal Energy Transfer in Mixtures Purpose: The purpose of this experiment is to discover how exactly the final temperature of a mixture, involving a substance and hot water, is affected and impacted by the type of substance used. This means that when hot water is mixed with another substance, it must be determined
Explanation:
did this on edge
What net force is needed to increase the velocity of a 5 kg object by 3 m/s over the course of 2.5 seconds?
Hi there!
Recall Newton's Second Law:
[tex]\large\boxed{\Sigma F = ma}[/tex]
∑F = net force (N)
m = mass (kg)
a = acceleration (m/s²)
We must begin by solving for the acceleration using the following:
a = Δv/t
In this instance:
Δv = 3 m/s
t = 2.5 sec
a = 3/2.5 = 1.2 m/s²
Now, plug this value along with the mass into the equation for net force:
[tex]\Sigma F = 5(1.2) = \boxed{6 N}}[/tex]
Answer:
[tex]\boxed {\boxed {\sf 6 \ Newtons}}[/tex]
Explanation:
According to Newton's Second Law of Motion, force is the product of mass and acceleration. The mass of the object is 5 kilograms, but the acceleration is unknown.
We can find acceleration using the following formula:
[tex]a= \frac{ \Delta v}{t}[/tex]
The velocity of the object increases by 3 meters per second. It accelerates in 2.5 seconds.
Δv= 3 m/s t= 2.5 sSubstitute the values into the formula.
[tex]a= \frac{3 \ m/s}{2.5 \ s}[/tex]
[tex]a= 1.2 \ m/s/s = 1.2 \ m/s^2[/tex]
Now we know the mass and acceleration.
m= 5 kg a= 1.2 m/s²Substitute the values into the force formula.
[tex]F=ma[/tex]
[tex]F= (5 \ kg)(1.2 \ m/s^2)[/tex]
[tex]F= 6 \ kg*m/s^2[/tex]
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. Our answer of 6 kilogram meters per second squared is equal to 6 Newtons.
[tex]F= 6 \ N[/tex]
a)iii) as the wheelbarrow was being pushed, the tyre went over a small stone. Describe and explain how the pressure changes.
Answer:
it changed because the rock put the pressure all in one spot.
Explanation:
The mass of a water skier is 54 kg. Calculate his/her weight if the gravitational field strength is 10 N/kg.
Answer:
weight=mass×gravity
the mass in this question is 54kg and the gravity is 10
therefore:
w=54×10
=540N
I hope this helps
Give 4 facts about the alkali metals
Answer:
there very old
Explanation:
The force involved in falling of an apple from a tree is known as? (a) Magnetic force (b) electrostatic force (c) Contact force (d) gravitational force
Answer:
gravitational force
...........
If you move two objects with opposite charges apart, what happens to the potential energy between them? Explain your response.
I have read it in the ncert science book
That any material charged with the same energy they repels each other while if they have opposite energy they attracts each other.
Hope it helps.
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A600kg car is acted force of 1200N starting from rest.how travel in 10 seconds?
Answer:
x= 100 m
Explanation:
m= 600 kg
t= 10 s
F= 1200N
v0= 0
-----
F=ma
a=F/m
x= (v0)t + (1/2)(a)(t^2)
x= (1/2)(F/m)(t^2)
x= (1/2)(1200/600)(10^2)
x= 100m
If mason runs a 5 k race at an average speed of 300 m/min how long will it take him to finish
[tex]\\ \sf\Rrightarrow Speed=\dfrac{Distance}{Time}[/tex]
[tex]\\ \sf\Rrightarrow Time=\dfrac{Distance}{Speed}[/tex]
[tex]\\ \sf\Rrightarrow Time=\dfrac{5000}{300}[/tex]
[tex]\\ \sf\Rrightarrow Time=16.6min[/tex]
In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.
Find the central force acting on the electron as it revolves in a circular orbit of radius 4.63 x 10-11 m.
Answer in units of N.
The central force acting on the electron as it revolves in a circular orbit is [tex]9.52 \times 10^{-8} \ N[/tex].
The given parameters;
speed of electron, v = 2.2 x 10⁶ m/sradius of the circle, r = 4.63 x 10⁻¹¹ mThe central force acting on the electron as it revolves in a circular orbit is calculated as follows;
[tex]F = \frac{M_e v^2}{r} \\\\[/tex]
where;
[tex]M_e[/tex] is mass of electron = 9.11 x 10⁻³¹ kg
[tex]F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N[/tex]
Thus, the central force acting on the electron as it revolves in a circular orbit is [tex]9.52 \times 10^{-8} \ N[/tex].
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A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.
a) What is the range of values of the initial velocity so that the projectile falls between points M and N?
Answer: A
Explanation:
I want my points so yea