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Basile [38]
2 years ago
7

In which graph is acceleration the slope?

Physics
2 answers:
shtirl [24]2 years ago
7 0

Answer:

it may help u I hope leave comment plz

olga2289 [7]2 years ago
4 0

Answer:

<em>Velocity</em><em> </em><em>-</em><em>time</em><em> </em><em>graph</em><em> </em>

Explanation:

hope it helps ✌️✌️

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Both retinal disparity and convergence increase as an object gets closer to the individual.
Gnom [1K]
That statement is true

Retinal disparity : space between your eyes that allow binocular vision to create depth perception

Retinal Convergence : Space between your eyes that signal visual moves to the retina

They both will increases as an object get closer to the individual, allowing them acknowledge and observe the existence of the object
4 0
3 years ago
Read 2 more answers
Which data set has the largest range?
MrRissso [65]

Answer:

Option C

Explanation:

We have to check range of all options first

For A:

Largest Value: 5

Smallest Value: 1

So range = Largest value - smallest value

5-1 = 4

For B:

Largest Value: 6

Smallest Value: 4

Range = 6-4 = 2

For C:

Largest Value: 9

Smallest Value: 1

Range = 9-1 = 8

For D:

Largest Value = 9

Smallest Value = 3

Range = 9-3=6

So, the data set in option C has the largest range

4 0
3 years ago
ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po
Levart [38]

Answer:

6.046N

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

F_n_e_t=F_g+F_t

F_t is the tension force.F_g=9.8N/kg is the force of gravity.

F_n_e_t=ma_c=mv^2/r\\

where r is the rope's radius from the fixed point.

From the net force equation above:

F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N

Hence the tension force is 6.046N

8 0
3 years ago
Question 1 of 10
Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

where k = 9*10^9Nm/C is the coulomb's constant.

Now, in our case

R = 0.0075m

Q = 0.0035C;

therefore,

E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}

\boxed{E = 5.6*10^{11}N/C.}

which is choice C from the options given<em> (at least it resembles it).</em>

6 0
3 years ago
An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of
Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

3\times \frac{\lambda}{2} = 0.39

\lambda = 0.26 m

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

d = \frac{\lambda}{4}

d = \frac{0.26}{4} = 0.065 m = 6.5 cm

So the distance will be 6.5 cm

3 0
3 years ago
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