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Alexandra [31]
3 years ago
15

which statements about acceleration are true? check all that apply. a. the si units of acceleration are m/s2. b. for acceleratio

n, you must have a number, a unit, and a direction. c. to calculate acceleration, divide the change in speed by the change in time. d. the symbol for acceleration is .

Physics
2 answers:
Salsk061 [2.6K]3 years ago
6 0
<span>A is True:

</span>The formula for acceleration is:

Acceleration = \frac{Vf-Vi}{t}
Acceleration = \frac{Xm/s-Xm/s}{s} = Xm/s^{2}

Because the unit of velocity is m/s and the unit of time is second this is what happened:

\frac{m}{s} ÷\frac{s}{1}

\frac{m}{s} ÷X \frac{1}{s} = \frac{m}{s^{2}} 

B is True:

Acceleration is a vector quantity, so that means it should have both magnitude and direction. When given an acceleration, the sign usually indicates the direction. A negative value is a sign that it is decelerating and not accelerating. 

D could be true only if:

Your question is actually broken so we do not know the symbol you are showing. But in general, the symbol of acceleration is a letter "a" with an arrow on top. You can see in the picture attached to see the symbol I am referring to. 

C is not true because acceleration is the change in VELOCITY over time. Velocity and speed are different. 



tatiyna3 years ago
5 0

The first thing you should know is that acceleration is a vector, therefore, it has magnitude, and direction.

By definition, we have that the vector acceleration is given by:

a = \frac{dv}{dt}\\

Where,

dv/dt: derived from the velocity vector with respect to time,

Therefore, the units of the vector acceleration are:

a = \frac{\frac{m}{s}}{s}\\

a = \frac{m}{s^2}

The symbol of acceleration is:

a^{->}

Answer:

The statements about acceleration that are true are:

a. the si units of acceleration are m/s2.

b. for acceleration, you must have a number, a unit, and a direction.

d. the symbol for acceleration is (True, only if the symbol is the same as above)

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Two forces whose resultant is 100N,are perpendicular to each other.if one of them makes an angle of 60° with the resultant, calc
DIA [1.3K]

Magnitude of the force = 50N

<u />

<u>Explanation:</u>

Let one force be X

And the second force be Y

Both the forces are perpendicular, so α = 90°

The resultant R of the two forces = 100N

Angle between resultant and 1 force, α = 60°

We know,

R² = X² + Y² + 2XYcos 90

100² = X² + Y²

If X makes an angle of Φ = 60°

Then, tan Ф =  Y sin α / X + Y cos α

Putting the values we find,

tan 60 = Y/X

Y  = √3X

Putting Y = √3X in equation 1 we get,

(X)² + (√3X)² = 100 X 100

4X² = 100 X 100

X² = 2500

X = 50N

Therefore, magnitude of the force = 50N

5 0
3 years ago
A 2.5 g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tre
Cloud [144]

Answer: Work done = 153.125Joules, Work done = 0.003Nm

Explanation:

Kinetic energy of a body is the energy possessed by a body by virtue of its motion.

Mathematically,

K.E = 1/2MV²

Where;

M = mass of the body = 2.5g = 0.0025kg

V = velocity of the body = 350m/s

Substituting this values in the formula, we have;

K.E = 1/2× 0.0025×350²

K.E = 153.125Joules

Work done is the force applied to body to cause it to move through a distance.

Work = Force × distance

Force = ma = 0.0025 × 10

Force = 0.025N

Distance = 12cm = 0.12m

Work = 0.025×0.12

Work = 0.003Nm

work done by the tree in stopping the bullet is 0.003N

4 0
3 years ago
Read 2 more answers
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an
Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

7 0
3 years ago
What is TeO3 compound name​
Stolb23 [73]

Tellurium trioxide

Hope this helps you!!!

4 0
3 years ago
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Can someone please help me with this
pentagon [3]
The key principle is that crank length, just like frame size, should be proportional to the rider height and then modified to what fits the individual. There are 4 charts, two for the upright position and two for the aero position, depending upon how you race.
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