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2 years ago

A velocity vs time graph what is the acceleration of the object

Physics

2 answers:

tigry1 [53]2 years ago

7 0

Answer:

If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s.

Explanation:

Dmitry [639]2 years ago

6 0

Answer:

The graph is missing, but ill try to answer this anyway.

The acceleration is the rate of change of the velocity; this means that if you know the velocity as a function of time; v(t), you can obtain the acceleration a(t) by differentiate v(t) with respect to time; this is:

a(t) = dv(t)/dt

Them if you have a graph of velocity versus time, you can find the equation of v(t) and then find the acceleration.

For example, if your graph of velocity vs time is a linear relation, then the acceleration will be the slope of the linear relation.

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A wave is incident on the surface of a mirror at an angle of 41° with the normal. What can you say about its angle of reflection

Oksana_A [137]

It's angle of reflection must be 41 degrees

we know, by the first law of reflection that angle of incidence is always equal to angle of reflection..........

4 0

2 years ago

Read 2 more answers

A parallel-plate capacitor is charged by connecting it across the terminals of a battery. If the battery remains connected and t

Tanya [424]

Answer:

The voltage ( or potential difference) V increases while the charge Q decreases.

Explanation:

The capacitance C of a capacitor is defined as the measure to which the capacitor can store charges. For a parallel-plate capacitor it is given by the following relationship;

$C=\frac{\epsilon_o\epsilon_rA}{d}............(1)$

where A is the surface area of the plates, d is their distance of separation, $\epsilon_o$ is the permittivity of free space and $\epsilon_r$ is relative permittivity.

Also, the capacitance of a capacitor can be expressed in the form of equation (2)

$C=\frac{Q}{V}...................(2)\\$

where Q is the charge stored and V is the potential difference.

By combining (1) and (2) and making d the subject of formula, we obtain the following;

$d=\frac{\epsilon_o\epsilon_rAV}{Q}............(3)$

By observing (3), it is seen that the distance d of separation between the plates is directly proportional to the potential difference V and inversely proportional to the charge stored Q. This implies that an increase in the distance d of separation will bring about an increase the the voltage or potential difference V and a decrease in the charge Q.

6 0

3 years ago

You are working in a biology lab and learning to use a new ultracentrifuge for blood tests. The specifications for the centrifug

Sloan [31]

Answer:

Yes, correct

Explanation:

velocity, v = 470 m/s

radius, r = 0.15 m

The radial acceleration is the centripetal acceleration which always acts towards the centre of the circular centrifuge.

The formula for the centripetal acceleration is given by

$a =\frac{v^{2}}{r}$

$a =\frac{470^{2}}{0.15}$

a = 1472666.667

a = 150272.1 g

According to the question, we can get the acceleration as mentioned. So the claim is correct.

7 0

2 years ago

Read 2 more answers

Consider a long cylindrical charge distribution of radius R with a uniform charge density rho. Find the electric field at distan

ahrayia [7]

Answer: Ok, first lest see out problem.

It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.

Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0

where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.

So Q= rho*volume= pi*r*r*L*rho

so replacing : E = (1/2)*r*rho/ε0

you may ask, ¿why dont use R on the solution?

since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.

R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.

3 0

2 years ago

The attractive electrostatic force between the point charges +8.44 ✕ 10-6 and q has a magnitude of 0.961 n when the separation b

d1i1m1o1n [39]

The electrostatic force between two charges Q1 and q is given by
$F=k_e \frac{Q_1 q}{r^2}$
where
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges

Re-arranging the formula, we have
$q= \frac{F r^2}{k_e Q_1}$
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
$q= \frac{(0.961 N)(0.67 m)^2}{(8.99 \cdot 10^9 N m^2 C^{-2})(+8.44\cdot 10^{-6}C)}=5.69 \cdot 10^{-6} C$

And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.

6 0

2 years ago