I need help with finding these?

A ball is hung from a rope, making a pendulum. When it is pulled 5° to the side, the restoring force is 1.0 N. What will be the

Trava [24]

Answer:

restoring force is 2 N

Explanation:

given data

angle pulled = 5°

force = 1 N

pulled = 10°

to find out

restoring force

solution

we know here force

force = m×g×sinθ ..........1

so here θ is very small so sinθ = θ

1 = mg(5)

mg = 1 /5 ..................2

and

now for 10 degree

we know here θ is small so sinθ = θ

so from equation 1

force = m×g×sinθ

put equation 2 here

force = 1/5 × 10

force = 2 N

so restoring force is 2 N

7 0

3 years ago

Describing Uses ñ Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. D

LiRa [457]

I don’t agree what if it blows up?

6 0

3 years ago

Please help me i need it it’s the only way i’ll pass

Varvara68 [4.7K]

Answer:

c. they travel through all forms of matter

Explanation:

3 0

3 years ago

A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must

Sergio039 [100]

Explanation:

The centripetal force $F_c$ on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

$y:\:\:\:\:N - mg = 0$

$x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N$

or

$m \dfrac{v^2}{r} = \mu mg$

Solving for $\mu$ ,

$\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28$

3 0

3 years ago

Two blocks are connected by a massless rope over a massless, frictionless pulley, as shown in the figure. The mass of block 2 is

Agata [3.3K]

The mass of block 1 is 1.30176 kg.

<h3>What is tension in physics?</h3>

A tension force in physics is a force developed in a rope, string, or cable when stretched under an applied force.

Tension is acted along the length of the rope/cable in a direction that is opposite to the force applied on it.

According to the question,

Mass of block 2= 11.3 kg

Kinetic friction μ=0.200

The component of the weight in the direction of the incline

W = m₂gsinθ

T = $F_{f} - W$ ........(1)

When the force of friction

$F_{f}$ = μ $_k$ N

$F_{f}$ = μ $_k$ (m₂gcosθ)

Here , Put the value in equation 1

T = m₂gcos 31.5° - μ $_k$ m₂gcos 31.5°

We know that ,

T - m₁g = 0

T = m₁g

So,

m₁g = m₂g (Sin 31.5° - 0.200 x cos 31.5°)

m₁ = 11.3 x (0.0840 - 0.2 x 0.996)

m₁ = 11.3 x (0.1152)

m₁ = 1.30176 kg

Hence, the mass of block 1 is 1.30176 kg.

Learn more about tension here:brainly.com/question/14294135

#SPJ1

7 0

2 years ago