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2 years ago

A force of 8,480 N is applied to a cart to accelerate it at a rate of 26.5 m/s2. What is the mass of the cart?

Physics

2 answers:

Anni [7]2 years ago

8 0

F=ma
8480=26.5m
m=8480/26.5
m=320
The mass of the cart is 320kg.

velikii [3]2 years ago

4 0

For the given question, the mass of the cart is 320 Kg.

Answer: Option B

<u>Explanation:</u>

According to Newton’s second law, the rate of change in momentum exerted by an object is directly proportional to the unbalanced force exerted on the object. So

$\text {Unbalanced Force } F=\frac{d(\text {momentum})}{d t}$

As momentum of an object is derived from the product of mass and velocity of the object. Thus,

$\text {Momentum}=m \times v$

So,

$\text {Unbalanced Force } F=\frac{d(m \times v)}{d t}$

As mass of an object tend to be same irrespective of time and velocity can vary with time, we obtain,

$\text {Unbalanced Force } F=m \times \frac{d v}{d t}=m \times \text { accelearation }$

As it is known that rate of change in velocity is equal to the acceleration exerted by the object. Thus, Newton’s second law can also be stated as that the external unbalanced force exerted on the object will be equal to the mass of the object and the acceleration exerted by the object. So in the present case, as we known the force and acceleration, the mass of the object will be determined by the formula F = ma.

$F=m \times \text {acceleration}$

Therefore,

$\text {Mass}=\frac{\text { Force }}{\text { Acceleration }}=\frac{8480}{26.5}=320 \mathrm{kg}$

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As light from a star spreads out and weakens, do gaps form between the photons?<br>

Zolol [24]

Answer:

There are no gaps in space between the photons as they travel. If you were to look at a wave then you'd come to a conclusion that indeed that there aren't any gaps unless they are specifically placed.The light from a distance star indeed spreads out and weakens as it travels, but this just reduces the wave strength and does not introduce gaps.

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2 years ago

A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?

Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

5 0

2 years ago

A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers

swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0

2 years ago

Read 2 more answers

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

4 0

2 years ago

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0

2 years ago