diagonal line means the velocity is constant.
Question: A) A car is driving too fast on a flat non-banked curve. The car cannot stay on the road. What path will the car take as it leaves the road? See Image. (a) (b) B) The driver wants to go faster around the same curve mentioned above and not run off the road. He adds sandbags to his car to make it weigh more so that the friction force between the tires and the road will be increased. Will this work? Explain your answer.
Explanation:
(a) The car will leave tangentially to the road, so it will take path d.
(b) Sum of forces on the car in the vertical direction:
∑F = ma
N − mg = 0
N = mg
Sum of forces on the car in the centripetal direction:
∑F = ma
Nμ = m v²/r
mgμ = m v²/r
gμ = v²/r
v = √(grμ)
The maximum velocity is independent of mass, so adding sandbags will not work.
Calculate the velocity of the bicyclist between 0 and 3 seconds.
Answer:
is there an equasion it gives you?
Explanation:
need a little more info.
How many Newtons of force does it take to move a 53 kilogram object?
What are the two main types of star clusters?
Answer:
Open and globular
Explanation:
An object is released from rest from a top of a building 90 meters high. Neglect air friction.
What is the volume of its acceleration?
Calculate the time it takes to reach the floor?
With what velocity does it reach the floor?
How fast is it moving when it is 50 meter above the floor?
Answer:
1. a = 9.8 m/s²
2. t = 4.28 s
3. Vf = 42 m/s
4. Vf = 28 m/s
Explanation:
1.
Since, the body is under free fall motion. Therefore, the value of its acceleration shall be equal to the acceleration due to gravity.
a = 9.8 m/s²
2.
The time taken by the ball to reach the ground can be calculated by using second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height = 90 m
Vi = initial velocity = 0 m/s
t = time taken = ?
Therefore,
90 m = (0 m/s)t + (1/2)(9.8 m/s²)t²
t = √(18.36 s²)
t = 4.28 s
3.
In order to find final velocity we use first equation of motion:
Vf = Vi + gt
Vf = 0 m/s + (9.8 m/s²)(4.28 s)
Vf = 42 m/s
4.
when the ball is at height of 50 m, it means it has covered:
h = 90 m - 50 m = 40 m
we use third equation of motion at this point:
2gh = Vf² - Vi²
(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²
Vf = √(784 m²/s²)
Vf = 28 m/s
You stand at the top of a cliff while your friend stands on the ground below you. You drop a ball from rest and see that it takes 1.7s for the ball to hit the ground below. Your friend then picks up the ball and throws it up to you, such that it just comes to rest in your hand. What is the speed with which your friend threw the ball
Answer:
16.7m/s
Explanation:
Using v= u+ at
V = final velocity. U = initial velocity
So since the two movements are the same they will return with the same speed they went up
Then final velocity v= 0
So
0:= u +-9.8 x 1.7
U= 16.7m/s
Question 3
A box is being pulled by a rope that makes a 25 degree angle with the ground. The
pulling force is 100.0 N
along the rope. Find the horizontal and vertical components of the force vector.
A. 90.63 N, 42.26 N
B. 86.87 N, 32.17 N
C. 60.87 N, 75,63 N
D. 80.9 N, 45.5 N
What is the answer A,B,C or D?
Answer:
A
Explanation:
The mass of a string is 20 g and it has a length of 3.2 m. Assuming that the tension in the string is 2.5 N, what will be the wavelength of a travelling wave that is created by a sinusoidal excitation of this string with a frequency of 20 Hz. Provide the wavelength in units of m. Please note: You do not include the units in your answer. Just write in the number.
Answer:
The wavelength of the wave is 1 m
Explanation:
Given;
mass of the string, m = 20 g = 0.02 kg
length of the string, L = 3.2 m
tension on the string, T = 2.5 N
the frequency of the wave, f = 20 Hz
The velocity of the wave is given by;
[tex]v = \sqrt\frac{T}{\mu} {}[/tex]
where;
μ is mass per unit length = 0.02 kg / 3.2 m
μ = 6.25 x 10⁻³ kg/m
[tex]v = \sqrt{\frac{T}{\mu} } \\\\v = \sqrt{\frac{2.5}{6.25*10^{-3}} } \\\\v = 20 \ m/s[/tex]
The wavelength of the wave is given by;
λ = v / f
λ = (20 m/s )/ (20 Hz)
λ = 1 m
Therefore, the wavelength of the wave is 1 m
blutions math practice
Maddie Hicks: Attempt 1
07.5 ml
Question 4 (1 point)
You have a stock bottle of 3mM tetrodotoxin solution, but your protocol states that
you need 6ml of 1mM tetrodotoxin (MW=319.27 g/mol). How much of your 3mM
stock solution (liquid) should you use to make the 1mM solution?
Answer:
[tex]V_1=2mL[/tex]
Explanation:
Hello,
In this case, considering this as a dilution problem, the first step here is to consider that the moles of tetrodotoxin remains the same and just the volumes and concentrations are modified from the initial stock (1) and the required dilution (2):
[tex]V_1M_1=V_2M_2[/tex]
Whereas V is referred to volume and M to molar concentration, in this case in mM. In such a way, solving the V1 as the volume of the 3-mM solution we obtain:
[tex]V_1=\frac{V_2M_2}{M_1} \\\\V_1=\frac{6mL*1mM}{3mM} \\\\V_1=2mL[/tex]
It means you need 2 mL of the 3-mM solution.
Best regards.
What is the speed of an object that travels 20 meters in 4 seconds?
Given:-
Distance covered by the particle (s) = 20 mTime taken (t) = 4 secondsTo Find: Speed of it.
We know,
v = s/t
where,
v = Speed,s = Distance &t = Time taken.Putting the values,
v = (20 m)/(4 s)
→ v = 5 m/s (Ans.)
The speed of an object will be "5 m/s".
The given values are:
Distance,
d = 20 mTime,
t = 4 secondsAs we know, the formula:
→ [tex]Distance (d)= Speed (v)\times Time (t)[/tex]
or,
→ [tex]Speed (v) = \frac{Distance (d)}{Time (t)}[/tex]
By putting the values, we get
→ [tex]= \frac{20}{4}[/tex]
→ [tex]= 5 \ m/s[/tex]
Thus the above answer is correct.
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it is required that an object weighing 8N be moved up ward at an acceleration of 2m/s what is the upward force needed
Answer:
F = 1.6 N
Explanation:
Newton's Second Law of Motion states that whenever an unbalanced force is applied on a body. It produces an acceleration in the body, in its own direction. The magnitude of the force is given by the following formula:
F = ma
where,
F = Force Required to Move the Object Upward = ?
m = mass of object
a = upward acceleration = 2 m/s²
but, we know that weight of object is given as:
Weight = mg
8 N = m(9.8 m/s²)
m = 0.8 kg
using values in the equation:
F = (0.8 kg)(2 m/s²)
F = 1.6 N
Help please Subject (general psychology)
Answer:
plz give the question... floor the proper answer
Explanation:
plz mark me add the brainliest
thank
me and follow ma
A person drops two objects from the same height. One object weighs 15 N.
and the other weighs 10 N. How does the mass of the objects relate to the
force of gravity on them?
A. The 10 N object has the same mass as the 15 N object.
B. The 15 N object has twice the mass of the 10 N object.
C. The 15 N object has more mass than the 10 N object.
D. The 10 N object has more mass than the 15 N object.
Answer:
C
Explanation:
The 15N object has more mass than the 10N object.
A 15 N object has more mass than the 10 N object due to their weight differences.
What is Weight?
Weight is defined as the force exerted by the earth on an object on the surface of earth. The direction of this force is directed towards the center of earth. Mathematically -
W = mg
Where -
m is the mass of body
g is the acceleration due to gravity
Weight can also be written as -
W = ρVg [ρ is density and V is volume]
Given is a person who dropped two objects from the same height. One object weighs 15 N and the other weighs 10 N.
We can write -
Weight of Object 1 [W1] = 15 N
Weight of Object 2 [W2] = 10N
We know that -
W = m x g
We can write it as -
m = W/g
Therefore -
Mass of Object 1 [M1] = 15/9.8 = 1.53 Kg
Mass of Object 2 [M2] = 10/9.8 = 1.02 Kg
Clearly -
M1 > M2
So, the 15 N object has more mass than the 10 N object.
Therefore, a 15 N object has more mass than the 10 N object due to their weight differences.
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A car moves from rest with an acceleration of 0.2ms
Find its velocity when it has moved a
distance of 50m
From the problem we notice that the question
does not involver. Therefore, the required equation
is)
Explanation:
Given:
Δx = 50 m
v₀ = 0 m/s
a = 0.2 m/s²
Find: v
The problem does not involve time. Therefore, the required equation is:
v² = v₀² + 2aΔx
Plug in values and solve:
v² = (0 m/s)² + 2 (0.2 m/s²) (50 m)
v = 4.47 m/s
The velocity of the car when it has moved a distance of 50 meters is approximately 4.47 m/s.
To find the velocity of the car when it has moved a distance of 50 meters, we can use the following equation of motion:
v² = u² + 2as
Where:
v = final velocity
u = initial velocity (which is 0 in this case as the car starts from rest)
a = acceleration
s = distance traveled
Plugging in the values into the equation:
v² = 0² + 2(0.2)(50)
v² = 0 + 20
v² = 20
Taking the square root of both sides:
v = √20
v = 4.47 m/s
Therefore, the velocity of the car when it has moved a distance of 50 meters is approximately 4.47 m/s.
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11. एक समान चुम्बकीय क्षेत्र में लम्बवत प्रवेश करने वाले किसी आवेशित कण द्वारा प्राप्त वृत्तीय पथ की
7 त्रिज्या का सूत्र ज्ञात कीजिए।
2
Find the formula of radius obtained by a charged particle entering perpendicularly
in a uniform magnetic field.
Answer:
r=mv/BeExplanation:
If a positive charge enters a magnetic field at 90 degrees the charge is deflected in a circular path by a force that acts perpendicular to it in line with Flemings right-hand rule
to derive the radius of the path of the charge we apply
F= mv^2/r=Bev
where
m= mass of the electronic charge
e=charge
B=magnetic field
v=average speed
r=radius
rearranging we have
r=mv^2/Bev
r=mv/Be
Question
In order for work to be done, what three things are necessary
Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 108/ (this is the speed of light in vacuum). How far (in meters) into space did the signal travel during the first 10 minutes?
Answer:
18*10^10 meters
Explanation:
V= d/t 10 mins = 600 seconds
3*10^8 = d/600s
(3*10^8)*(6*10^2) = d
d = 18*10^10 m
Which kinds of objects emit visible light in the electromagnetic spectrum? A. all objects B. radioactive objects C. relatively cold objects D. relatively hot objects
Answer:
c
Explanation:
Answer:
the answer is c i just did it
Explanation:
The speed of x-rays is 300 000. m/s. In scientific notation, and with the correct number of significant figures, this speed is… *
Answer:
3.0 x 10 and the exponent on ten is 5.
Explanation:
You are moving the decimal point over to the left 5 times, making the exponent a positive 5 and then you would put a decimal after 3. Making it 3.0 x 10, and the exponent 5.
The speed of x-rays in scientific notation is 3.0 x 10⁶ m/s.
The given parameters;
speed of the x-ray, v = 300,000 m/s
The standard form of the given number represents the scientific notation of the number.
To represent a number in standard form, multiply the number in terms of power or exponent of 10. The value of the exponent represents the number of available zeros present.In scientific notation the speed of the x-ray can be expressed as follows;
300,000 m/s = 3.0 x 10⁶ m/s
Thus, the speed of x-rays in scientific notation is 3.0 x 10⁶ m/s.
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Which phrase describes a scientific law?
A. A statement that matter cannot be created or destroyed
B. A claim that experiments cannot verify whether matter has been
destroyed
O C. An explanation for why matter cannot be created or destroyed
D. A prediction of how much matter exists in the universe
SUBMIT
In
Answer:
C- An explanation for why matter cannot be created or destroyed
Explanation:
Answer: A. A statement that matter cannot be created or destroyed.
Explanation:
Apex
Elmer Fudd is walking at 2.0 m/s. If he walks for 2.0 minutes, how far does he walk?
(Can someone help with the answer due in 5min)
Answer:
240 meters
Explanation:
He is walking 2 meters per second, and there are 60 seconds per minute. There are 120 seconds in two minutes. So, 120 x 2 = 240. That's your answer!
If you walk 5 km north and then 12 km east. Your resultant displacement is____.a. magnitude: 1.1km; direction: 53.1 degrees east of north.b. magnitude: 2.00; direction: 53.1 degrees east of north.c. magnitude: 2.00; direction: 36.9 degrees east of north.d. magnitude: 1.1km; direction: 36.9 degrees east of north.
Answer:
Resultant = 13km
Direction = 67.38° East of North
Explanation:
Given the following :
5km North ; 12km East
Resultant Displacement (r) :
r² = 5² + 12²
r² = 25 + 144
r² = 169
r = √169
r = 13
Direction:
Tangent = opposite / Adjacent
Tanθ = opposite / Adjacent
Opposite = 12 ; adjacent = 5
Tanθ = (12/5)
Tanθ = 2.4
θ = tan^-1(2.4)
θ = 67.38° east of north
Do states give up all power to withdraw when they join a union of states, or do they still retain some control over their fate?
Calculate the length of segment RS with midpoint, M , if RM = 5x and MS = x + 12.
A. 3
B. 25
C. 30
D. 36
E. 15
Answer:
the full segment is: 30 units long
which coincides with answer C in your list
Explanation:
If M is the midpoint, then it divides the segment in two equal parts. Then, we can say that:
RM = MS (equality among the two parts of the divided segment)
replacing RM with "5 x" and MS with "x + 12", we get:
5 x = x + 12
solving for x:
5 x - x = 12
4 x = 12
then x = 3
With this info, we can calculate the length of each half of the segment and consequently its full length:
RM = 5 x = 5 (3) = 15
then the full segment is: 30 units long
Direction of propagation of light is:________a. in all directions irrespective of polarization b. parallel to the polarization c. perpendicular to the polarization d. not dependent on polarization
Answer:
Perpendicular to the polarization
Explanation:
In this question, we are interested in selecting the option that best complete the question.
The direction of propagation of light is perpendicular to the polarization. Polarized and unpolarized light behave in different manners.
In unpolarized light, the oscillation of the magnetic and electric occurs in all directions to the propagated light.
However, in the case of polarized light, there is a restriction of the electric field oscillation. And thus, it only oscillates in a direction which is perpendicular to the propagation light
In which direction does centripetal force act on an object? in the opposite direction of the tangential speed of the object in the same direction of the tangential speed of the object toward the outside of the circle in which the object is moving toward the center of the circle in which the object is moving
Answer:
towards the center of the circle
(which appears to be the last option in your list of possible answers)
Explanation:
The centripetal force points always towards the center of the circle described by the object moving.
The direction of centripetal force acting on an object is always towards the center of the circle in which the object is moving.
The given problem is based on the concept and fundamentals of centripetal force. When an object is known to move around a circular path, then there is a force comes to play acting towards the centre of circular path, known as centripetal force.
In other words, the centripetal force is also known termed as the force that tends to act on a body moving in a circular path and is directed towards the centre around which the body is moving.For example - Motion of merry go round and the motion of roller skates on the rink floor are the examples of centripetal force, where the force always acts in the direction, center of the circular track.Thus, we can conclude that the direction of centripetal force acting on an object is always toward the center of the circle in which the object is moving.
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The mere exposure effect is
65°F to degrees Celsius
Answer:
18.3 C
Explanation:
Answer: 55/3 °C
Explanation:
Formula:
C=5/9(F-32)
Given:
F=65
Solve:
C=5/9(F-32)
C=5/9(65-32)
C=5/9(33)
C=55/3
How does deploying slats or slots on an airfoil affect CL max and Stall AOA?
a) Stall AOA increases, CL max increases.
b) Stall AOA decreases, CL max increases.
c) Stall AOA increases, CL max decreases.
d) Stall AOA decreases, CL max decreases.
Answer: a. Stall AOA increases, CL max increases.
Explanation:
Deploying slats or slots on an airfoil affect CL max and Stall AOA by increasing Stall AOA and also increasing CL max.
It should be noted that the coefficient of lift CL would rise through the use of slots due to increase in boundary later energy. The slots also leads to delay of stall by through increase in AOA.
How much work must be done to stop a 1150 kg car traveling at 120 km/h?
Answer:
The work to be done is 638,761.12 J
Explanation:
Given;
mass of the car, m = 1150 kg
velocity of the car, v = 120 km/h = 33.33 m/s
Apply work energy theorem;
W = K.E
where;
W is the work done
K.E is the kinetic energy
W = ¹/₂mv²
W = ¹/₂ (1150)(33.33)²
W = 638,761.12 J
Therefore, the work to be done is 638,761.12 J