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GREYUIT [131]
3 years ago
7

Several forces act on an object at rest. It is known that the sum of the forces acting on the object is zero. Which statement is

necessarily true?
Question 7 options:

The object's center of mass may move in such a way that the object will roll without slipping.

The object's center of mass may accelerate, but the object will remain in the same orientation.

The object's center of mass may accelerate and the object may begin to rotate.

The object will remain stationary.

The object's center of mass will not accelerate, but the object may begin to rotate.
Physics
1 answer:
MAVERICK [17]3 years ago
4 0

Answer:

When net force of zero acting on a ball which is at rest , then the object center of mass will not accelerate , but the object may begin to rotate .

Explanation:

Here when there is an object where several forces are acing upon  are zero then the center of mass will not accelerate because we know that

    A_{com}=\frac{F_{net}}{M}

Where A_{com} = acceleration of center of mass  

            F_{net} = net force = 0

   So the acceleration of center of mass will be zero

But the torque ,may not be zero as torque is product of individual force and perpendicular force .

 Since if torque is not equal to zero then the object may begin to rotate

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A rocket at fired straight up from rest with a net upward acceleration of 20 m/s2 starting from the ground. After 4.0 s, the thr
8_murik_8 [283]

Answer:

The maximum height reached by the rocket is 486.53 m

Explanation:

Given;

initial velocity of the rocket, u = 0

acceleration of the rocket, a= 20 m/s²

duration of the rocket first motion, t = 4 s

The distance traveled by the rocket before its thrust failed

h₁ = ut + ¹/₂at²

h₁ = 0 + ¹/₂ x 20 x 4²

h₁ = 160 m

The second distance moved by the rocket is calculated as follows;

The velocity of the rocket before its thrust failed;

v = u + at

v = 0 +  20 x 4

v = 80 m/s

This becomes the initial velocity for the second stage

At maximum height, the final velocity = 0

v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m

The maximum height reached by the rocket = h₁ + h₂

                                                                          = 160 + 326.53

                                                                          = 486.53 m

6 0
2 years ago
what will happen to the fieldof view for each resultant magnification as you change objectives from 4 to 10 to 43
jonny [76]

Answer:

The field of view is reduced.

Explanation:

Given that,

The field of view for every resultant magnification like you change objectives from 4 to 10 to 43.

We know that,

Field of view :

When the view is observed at a point in a defined field then these field called field of view.

The normal angle of  field of view is 90°.

The formula of field of view is define as,  

field\ of\ view = \dfrac{field\ number}{magnification}

We can say that,

The field of view is inversely proportional to the magnification.

When magnification is low then field of view will be large.

When magnification is higher then field of view will be small .

According to question,  

When the magnification adjust from 4 to 10 to 43, the field of view is reduced.    

Hence, The field of view is reduced.

7 0
3 years ago
A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total
worty [1.4K]

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

6 0
3 years ago
You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-
Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

E_P=\int dE \cos

E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}

\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ

\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}

If we put an electron on point P, then force on point e is :

\vec{F}=-|e|\vec{E_P}

F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}

If r >> x , then    $\frac{x^2}{r^2} \approx 0$

Then,  $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0
3 years ago
Give an example of a reputable website
Leni [432]
Well, almost any website can be, but if you are meaning for money, then google. if you are meaning reliable, then look for something with .org, its history and stuff
6 0
3 years ago
Read 2 more answers
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