All categories

3 years ago

5. Gravitational and magnetic forces are examples of _______forces.

Physics

1 answer:

Anvisha [2.4K]3 years ago

8 0

Answer:

Explanation:

5.non contact force

6.balanced force

7.unbalanced force

8.net force

9.zero

10.is the difference between two forces

mark me as brainliest please

You might be interested in

What are 2 things Lithium, Beryllium, and Boron, have in common?

ivann1987 [24]

Answer:

larger atomic radii because they are on the far left, the s-block, of the periodic table. -The ionic radii increase as you go down groups on the periodic table. -Lithium and Beryllium are not examples of this trend because they are both in 2s.

8 0

3 years ago

Help a girl out ?! Plz ill give brainlest

shepuryov [24]

Answer:

I am pretty sure it is B.

Explanation:

I hope this helped if it didn't I am truly sorry

5 0

3 years ago

Read 2 more answers

When is the magnitude of the acceleration of a mass on a spring at its maximum value?

DiKsa [7]

Answer:

Explanation:

Now we know that Force is the rate of change of momentum meaning

F= mv/t

But

mv/t = Ke

v/t = ke/m

a= ke/m

Where a is acceleration

K is constant of proportionality of tension on a spring

e is the extension substended by a string.

From the formula of acceleration we can see that as mass decreases acceleration increases so we can see that m = 1

We would have a maximum value of acceleration.

6 0

3 years ago

Examine the scenario.

frutty [35]

I Think The answer is d I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you

6 0

3 years ago

Read 2 more answers

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

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

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

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

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

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

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

4 years ago