Answer:
The car is considered to be speeding
Explanation:
The following data were obtained from the question:
Constant acceleration (a) = 7.4 m/s²
Time (t) = 2.4 s
Speed limit = 15.64 m/s
To know if the car speeding at any point, we shall determine the velocity of the car.
This can be obtained as follow:
Constant acceleration (a) = 7.4 m/s²
Time (t) = 2.4 s
Constant velocity (v) =?
a = v/t
7.4 = v / 2.4
Cross multiply
v = 7.4 × 2.4
v = 17.76 m/s
Thus, the velocity of the car is 17.76 m/s.
Comparing the velocity of the car (i.e 17.76 m/s) with the speed limit (i.e 15.64 m/s), we can see that the velocity of the car is higher than the speed limit. Therefore, the car is considered to be speeding.
This cantilever beam is loaded by two point loads. If the moment of inertia of the beam is 5.0*10-6 m4 and the modulus of elasticity is 210 GPa, the maximum deflection of the beam is:
Answer:
The answer is i just need the 5 pt sowwry bro hope you have luck finding it ;]
Explanation:
you prolly alr found it but it doesnt really matter
An 85 kg man stands in a very strong wind moving at 15 m/s at torso height. As you know, he will need to lean in to the wind, and we can model the situation to see why. Assume that the man has a mass of 85 kg, with a center of gravity 1.0 m above the ground. The action of the wind on his torso, which we approximate as a cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground, produces a force that tries to tip him over backward. To keep from falling over, he must lean forward.
Required:
a. What is the magnitude of the torque provided by the wind force? Take the pivot point at his feet. Assume that he is standing vertically.
b. At what angle to the vertical must the man lean to provide a gravitational torque that is equal to this torque due to the wind force?
Answer:
a) [tex]T= 72.9 Nm[/tex]
b) [tex]\Theta = 5 \textdegree[/tex]
Explanation:
From the question we are told that
Mass 85kg
Speed 15m/s
Center of gravity 1.0m
Cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground
a)Generally equation for force is given by
[tex]F = 0.5 p A v^2[/tex]
Mathematically solving for force exacted
[tex]F = 0.5*1.2*0.9*0.5*15^2[/tex]
[tex]F= 60.75 N[/tex]
Mathematically solving for torque
Torque,
[tex]T = r * F[/tex]
[tex]T=1.2*60.75[/tex]
[tex]T= 72.9 Nm[/tex]
b)Generally in solving for [tex]\theta[/tex]
[tex]Tan\theta = Torque/(Mass * Gravity)[/tex]
[tex]Tan\theta = (72.9)/(85 * 9.8)[/tex]
[tex]\Theta = 5 \textdegree[/tex]
a tennis ball rolls horizontally to the left for 30 m at a constant speed. what is the velocity after 15 seconds? what is the speed?
**PLEASE HELP - for physics, list variables
Q2. You push a crate up a ramp with a force of 10 N. Despite your pushing, the crate slides down the ramp 4 m. How much work did you do
Answer:
40 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 10 N
Distance (s) = 4 m
Workdone (Wd) =?
Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:
Workdone = force × distance
Wd = F × s
With the above formula, we can obtain the workdone as follow:
Force (F) = 10 N
Distance (s) = 4 m
Workdone (Wd) =?
Wd = F × s
Wd = 10 × 4
Wd = 40 J
Thus, 40 J of work was done.
Question 3 (1 point)
here were 2cars racing a quarter mile. The green car had a mass of 1200kg and crossed the finish line with a velocity of 53m/s. The red car had
a mass of 1100Kg and crossed the finish line with a velocity of 55m/s.,Which car had the great momentum?
Black car
Blue car
Green car
Red car
Answer:
The green car had the greatest momentum
Explanation:
Momentum
Momentum can be defined as "mass in motion" and is calculated as the product of the mass of the object by its velocity.
Being v the magnitude of the velocity and m the mass of the object, the momentum is calculated with:
p = mv
The green car had a mass of m1=1200 kg and crossed the finish line at v1=53 m/s. Hence, its momentum was:
p1 = 1200 Kg * 53 m/s = 63600 Kg.m/s
The red car had a mass of m2=1100 kg and crossed the finish line at v2=55 m/s. Hence, its momentum was:
p2 = 1100 Kg * 55 m/s = 60500 Kg.m/s
Since p1 > p2, then the green car had the greatest momentum
An inductor with an inductance of .5 henrys (H) is to be connected to a 60 Hz circuit. What will the inductive reactance (X L) be
Answer:
1885.2 ohms
Explanation:
Step one:
given data
L=5H
f=60Hz
Required
The inductive reactance of the inductor
Step two:
Applying the expression
XL= 2πfL
substitute
XL=2*3.142*60*5
XL=1885.2 ohms
As an object’s speed decreases, its kinetic energy (KE) ________.
As an object’s height increases, its gravitational potential energy (PEg) ________.
What purpose of government is shown in the First Amendment?
They guarantee rights of individuals.
They help people cooperate
They limit the rights of individuals
They provide services
Answer:
i think its They Guarantee rights of individuals
Explanation:
Answer:
To guarantee freedoms
Explanation:
Took the test
Does a basketball, baseball, tennis ball, or marble MOST LIKELY have the smallest volume?
Answer: tennis ball
Explanation:
A 0.41kg football that is initially at rest acquires a velocity of 35m/s when it is kicked. If the kicker's boot remains in contact with the ball for 0.08s, what is the average force of the kick?
Answer:
F = 318.88[N]
Explanation:
This problem can be solved by the principle of momentum conservation, which tells us that momentum is preserved before and after kicking the ball.
In this way, we can construct the following equation.
[tex]F*t=m*v[/tex]
where:
F = force [N]
t = time = 0.08 [s]
m = mass = 0.41 [kg]
v = velocity [m/s]
[tex]F*0.045=0.41*35\\F=318.88[N][/tex]
Food Chain: is a sequence of feeding relationships which describes which organism eats another.
● The Direction of the arrow in a food chain means:_____________________
Answer:
Naruto usamaki is the greatest hokagey in the leaf village
Explanation:
I took the test on edge
What do C-microwaves reveal to scientists?
Answer:
Different wavelengths of microwaves (grouped into "sub-bands") provide different information to scientists. Medium-length (C-band) microwaves penetrate through clouds, dust, smoke, snow, and rain to reveal the Earth's surface.
A skateboarder traveling at 10.4 m/s starts getting chased by a grumpy dog. The skateboarder speeds up with a constant acceleration for 39m over a 3.3s. We want to find the acceleration of the skateboarder over the 3.3s time interval. Which kinematic formula would be most useful to solve for the target unknown?
a. V= Vo+a(t)
b. X= (V+Vo/2)t
c. X= Vo(t)+1/2at^2
d. V^2 = Vo^2 +2ax
e. X= vt-1/2at^2
Answer:
c)X= Vo(t)+1/2at^2
Explanation:
From the question we are told that
initial speed u=10.4
displacement d= 39
time t=3.3s
accelaration a=?
Generally the best equation in solving for Acceleration is given to be
X= Vo(t)+1/2at^2
Mathematically
[tex]X= Vo(t)+1/2at^2[/tex]
[tex]39=10.4t+\frac{a}{2} *3.3^2[/tex]
[tex]5.445a=39-34.32[/tex]
[tex]a=39-34.32/5.445[/tex]
[tex]a=4.68/5.445[/tex]
[tex]a=0.86m/s^2[/tex]
What are your ideas about why you cannot see three different colors?
Answer:
Because humans possess these three types of cones, they are described as trichromats (tri = 3, chroma = color). ... The perception of color is not simply a matter of wavelength detection, as it involves various stages of neuronal processing
A constant net force of 345 N is applied to slide a heavy stationary
couch across the floor. If the force is contact with the couch for 1.5 seconds,
then what is the change in momentum of the couch
Answer:
P = 517.5 [kg*m/s]
Explanation:
We must remember that momentum is defined as the product of force by the time of force duration.
[tex]P=F*t[/tex]
where:
P = momentum [kg*m/s]
F = force [N]
t = time [s]
[tex]P=345*1.5\\P=517.5[kg*m/s][/tex]
The metal wire is stretched so that its cross-section is still circular but its total length is now 10 meters. What is the resistance of the wire after stretching
HELP ASAP!
Everything on screenshot.
Answer:
11. D
12. A
13. B
Explanation:
Which type of physical activity is being performed in the picture?
A.
cardiovascular
B.
strength training
C.
flexibility
D.
cross training
Reset Next
fill in the blank:
the more people (greater mass) you load into a car, the ___ (more or less) it will accelerate with the same force applied by the engine.
the more people (greater mass) you load into a car, the more it will accelerate with the same force applied by the engine.
14. After finishing her homework, Sue climbs up a 5.00 m high flight of stairs to her bedroom
Find the magnitude of Sue's weight
and how much
work Sue does in climbing the stairs if she
has a mass of 50.0 kg? (4.90 x 2 N, 2450J)
Explanation:
Given parameters:
Height = 5m
Mass of Sue = 50kg
Unknown:
Magnitude of Sue's weight = ?
Work done by Sue = ?
Solution:
Weight is the vertical force exerted by a body in the presence of gravity.
Mathematically;
W = mg
m is the mass
g is the acceleration due to gravity = 9.8m/s²
Weight = 50 x 9.8 = 490N
Work done = Force x distance = weight x height
Work done = 490 x 5 = 2450J
A 47.5-g ball moves at 30.0 m/s. If its speed is measured to an accuracy of 0.20%, what is the minimum uncertainty in its position
Answer:
The minimum uncertainty in its position is 1.85 x 10⁻³² m.
Explanation:
Given;
mass of the ball, m = 47.5 g = 0.0475 kg
speed of the ball, v = 30 m/s
measuring accuracy of the speed, = 0.2% = 0.002
The uncertainty in measurement of momentum;
ΔP = mΔv
ΔP = (0.0475)(30 x 0.002)
ΔP = 2.85 x 10⁻³ kgm/s
The uncertainty in position is calculated as;
[tex]\delta x \geq \frac{h}{4\pi (\delta P)}[/tex]
where;
h is Planck's constant
[tex]\delta x \geq \frac{6.626 \ \times \ 10^{-34}}{4\pi (2.85 \ \times \ 10^{-3})} \\\\\delta x \geq 1.85 \ \times \ 10^{-32} \ m[/tex]
Thus, the minimum uncertainty in its position is 1.85 x 10⁻³² m.
Choose all the answers that apply. Force has _____.
efficiency
direction
time
magnitude
A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N
weight. If the diameter decreases to 0.210 cm, determine (i) the
engineering stress and strain at this load and (ii) the true stress and strain
at this load
Answer: the answer is 19cm
Explanation:
WILL GIVE BRAINLIEST TO THE CORRECT ANSWER
Estimate the net force needed to accelerate a 1000-kg car from rest.
AND
Estimate the weight of a 1000-kg car that is accelerating at 3m/s/s.
Answer:
See the answers below.
Explanation:
We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F =m*a
where:
F = force [N] (units of newtons)
m = mass = 1000 [kg]
a = acceleration = 3 [m/s²]
[tex]F = 1000*3\\F=3000[N][/tex]
And the weight of any body can be calculated by means of the mass product by gravitational acceleration.
[tex]W=m*g\\W=1000*9.81\\W=9810 [N][/tex]
Will give brainliest if correct
Please help
Answer:
85.45
Divide 4,700/55
So C.
Answer:
C.85
Explanation:
Use the equation:
v= fλ ( v=wave speed, f= frequency, λ= wavelength)
4700= 55×λ
λ=4700÷55 =85.45 =85(approximately)
A person lifts a heavy load to a vertical height of 2.0 m in 3 seconds. If he/she had done this more slowly in 6 seconds, the
work on the load would have been:
Four times as great
half as great
the same
twice as great
Answer:
If the heavy load had been lifted more slowly, the work done on the load would have been the same.
Explanation:
Work done on an object is given as;
W = Fd
where;
F is the force applied on the object
d is the displacement of the object
for the given question, the applied force on the load = mg (mass of the load multiplied by acceleration due to gravity).
Also, the displacement of the object = vertical height the load was lifted.
W = mgh
The work done on the load is independent of time.
Thus, if the heavy load had been lifted more slowly, the work done on the load would have been the same.
A person lifting a heavy load to a vertical height of 2.0 m in 3 seconds does the same work as if he/she lifts it in 6 s.
A person lifts a heavy load to a vertical height of 2.0 m in 3 seconds.
We want to compare the work done with the one that he/she would have done if the process had taken 6 seconds.
What is work?In physics, work (W) is the energy transferred to or from an object via the application of force (F) along a displacement (s).
W = F × s
Given the displacement is the same (2.0 m) and the force needed is also the same (weight of the object), the work is the same for both processes.
A person lifting a heavy load to a vertical height of 2.0 m in 3 seconds does the same work as if he/she lifts it in 6 s.
Learn more about work here: https://brainly.com/question/25064916
⚠️Please Help Me! I Need It Now!⚠️
Answer:
A. They are moving at 60 mph
Explanation:
When looking at the car on the track pass by, the people inside appear to be moving at the same speed as the car. So if the car is going 60 mph, the people instead will appear as though they are moving at 60 mph as well.
as the temperature of a gas in a solid container increases what happens to the pressure exerted by the gas?
local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength at which they are broadcasting. Wavelength
Answer:
λ = 437 m.
Explanation:
Since a radio wave is a electromagnetic type of wave, it propagates approximately at the same speed of light in vacuum, 3*10⁸ m/s.As in any wave, there exist a fixed relationship between the propagation speed, the wavelength and the frequency of the radio wave, as follows:[tex]c = \lambda* f (1)[/tex]
Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:[tex]\lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m (2)[/tex]
Identify the velocity-versus-time plot(s) that correspond to motion under a constant, non-zero acceleration.
Answer:
See explanation
Explanation:
The question is incomplete because the images were not attached but I will try to help you as much as possible.
Constant acceleration implies that the velocity increases uniformly with time.
The graph of constant acceleration is a straight line graph having a slope. The slope of the graph is constant at any point along the straight line.
The image attached shows a velocity-time graph depicting constant acceleration.