All categories

3 years ago

The acceleration of an object would increase if there was an increase in the

Physics

1 answer:

xxMikexx [17]3 years ago

4 0

As per Newton's II law we know that

$F_{net} = ma$

here we know that

$F_{net} = F_{ap} - F_f$

so here we will have

$a = \frac{F_{ap} - F_f}{m}$

so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.

So here correct answer will be

A) force on the object.

You might be interested in

If you lift weights faster than normal you have increased what?

egoroff_w [7]

I belive it would power I’m not 100% sure. If it’s not power then force

8 0

2 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

The free body diagram shows a box being pulled to the left up a 25-degree incline

Greeley [361]

The question is incomplete but still I answer to assume your thinking.
The picture is attached below!.
Here,
F is the force with which you pull up the incline.
N is the normal force.
w is the weight acting downward.
Axis are mentioned in the attached picture.
Concept:
You can see there is no movement of object in the y-direction that means acceleration is zero in y-direction, sum of all the forces in y-direction equal to zero.
According to newton second law,
∑ F = ma
As, acceleration is zero in y-direction, so right hand side is zero in the above equation.
∑ F = 0
N-wcosθ=0
N= m*g*cos25°
N= m*(9.8)*(0.9063)
N= 8.8817*m
By putting the value of mass(m)(not given in the question) you will get the answer.
Hopefully, this is the answer of your question.

5 0

3 years ago

Read 2 more answers

The odometer of a car changes from 1048 km to 1096 km in 40

Makovka662 [10]

Answer:

20m/s

Explanation:

it covers 20 metres in a second

3 0

2 years ago

A firecracker in a coconut blows the coconut into three pieces. two pieces of equal mass fly off south and west, perpendicular t

netineya [11]

Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
$m_{1} v_{1i} + m_{2} v_{2i} + m_{3} v_{3i} = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}$
Before the firecracker blows a coconut does not move, so left side is equal to 0:
$0 = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}$

We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
$v_{3f} = - \frac{m_{1} v_{1f} + m_{2} v_{2f} }{ m_{3} }$

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
$v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s$
This speed we can insert into formula for v3f:
$v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\ \\ v_{3f} = - \frac{2m*33.9 }{ 2m } \\ \\ v_{3f} = - 33.9m/s$

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.

8 0

3 years ago

Read 2 more answers