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2 years ago

8

You push a cart with mass 15 kg forward , giving it an acceleration of 3 m/s ^ 2 . How much force did you apply ? O A. 0.2N O B.

O 18N O D. 45 N

2 answers:

Alexandra [31]2 years ago

7 0

Answer: force = mass x accelegation

Explanation: 15x3 = 45 N

kvasek [131]2 years ago

7 0

Answer:

A

Explanation:

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A grating whose slits are 3.2x10^-4 cm apart produces a third-order fringe at a 25.°0 angle. What is the wavelength of light tha

Ber [7]

Answer:

The light used has a wavelenght of 4.51×10^-7 m.

Explanation:

let:

n be the order fringe

Ф be the angle that the light makes

d is the slit spacing of the grating

λ be the wavelength of the light

then, by Bragg's law:

n×λ = d×sin(Ф)

λ = d×sin(Ф)/n

λ = (3.2×10^-4 cm)×sin(25.0°)/3

= 4.51×10^-5 cm

≈ 4.51×10^-7 m

Therefore, the light used has a wavelenght of 4.51×10^-7 m.

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3 years ago

The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electro

shusha [124]

Answer:

$1.6\times 10^{-18} C$

Explanation:

The fundamental charge of a single electron is $1.6\times 10^{-19} C$ .

If there are 10 excess electrons, the net charge that would be measured should be 10 times the fundamental charge of a single electron:

$Q=nq_e\\Q= 10\times 1.6\times 10^{-19} C\\Q= 1.6 \times 10^{-18} C$

3 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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3 years ago

A sprinter moving at 10 m/s slows down at a rate of 1.4 m/s2. how fast is the runner moving after 4 seconds?

AURORKA [14]

V=u+at
v=10+(-1.4 (4))
=10-5.6
=4.4m/s

3 0

3 years ago

Read 2 more answers

Miles is camping in Glacier National Park. In the midst of a glacier canyon,

valentina_108 [34]

Answer:

t=1.623 sec

Explanation:

The distance traveled before the echo is had is:

$distance=2d, d=280\ m\\\\=280\times 2\\\\=560 \ m$

Given the speed of sound as v=345m/s, we use the speed equation to solve for t:

$v=\frac{d}{t}\\\\345\ m/s=\frac{560m}{t}\\\\t=\frac{560}{360}\\\\=1.623 \ s$

Hence, it takes 1.623 seconds to hear the echo.

8 0

3 years ago