When is the average velocity of an object equal to the instantaneous velocity?

A student pushes a 21-kg box initially at rest, horizontally along a frictionless surface for 10.0 m and then releases the box t

Marrrta [24]

Answer:v=3.08 m/s

Explanation:

Given

mass of student $m=21 kg$

distance moved $d=10 m$

Force applied $F=10 N$

acceleration of system during application of force is a

$a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2$

using $v^2-u^2=2 as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-0=2\times 0.476\times 10$

$v=\sqrt{9.52}$

$v=3.08 m/s$

6 0

3 years ago

Tuesday

irinina [24]

If a teacher had 48 red pens and the ratio of red to blue pens she owns is 6:1, she will have a total of 56 pens

Let the number of red pens be R and the number of blue pens be B.

R = 48

R / B = 6 / 1

48 / B = 6 / 1

B = 48 / 6

B = 8

Total number of pens = Number of red pens + Number of blue pens

Total number of pens = R + B

Total number of pens = 48 + 8

Total number of pens = 56 pens

Therefore, she has a total of 56 pens

To know more about ratio

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3 0

1 year ago

A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal

Dmitry [639]

The resultant force on the animal = Resultant mass * total acceleration
F = 0.2 * 2.5 to the right
F = 0.5 to the right.

As, girl exerting a force of 3.5 N & it's not mentioned that she is in right or left, so the force exerting by boy would be either:
3.5-0.5 = 3 OR 3.5+0.5 = 4

If boy exerting a greater force then, answer will be 4 N & if girl exerting a greater force the, answer will be 3 N

Hope this helps!

3 0

3 years ago

Read 2 more answers

What is the correct definition of nurture?

makkiz [27]

Answer:

(C) the fulfillment of physical, mental, emotional, and social needs

Explanation:

Nurture

Definition (A) - Care for and encourage the growth or development of

Definition (B) - The process of caring for and encouraging the growth or development of someone or something

The best definition in the answer list for the word "nurture" is answer (C) The fulfillment of physical, mental, emotional, and social needs

8 0

3 years ago

Read 2 more answers

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

2 years ago