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natta225 [31]
2 years ago
9

Please help! will give brainlest!!!!!!!!!!!!

Physics
1 answer:
eimsori [14]2 years ago
7 0

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

a=0

Therefore the net force is zero as well:

\sum F = 0

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

\sum F = F cos \theta - F_f = 0

where

F cos \theta is the horizontal component of the applied force, with

F = 22.5 N

\theta=32.0^{\circ}

F_f is the force of friction

And solving for F_f, we find

F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

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A 38 kg crate rests on a floor. A horizontal pulling force of 170 N is needed to start the crate
Mandarinka [93]

Answer:

0.456033049

Explanation:

F=\mu N where N=mg hence

F=\mu mg where m is mass of object, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2}, \mu is the coefficient of static friction and F is the applied force.

Making \mu the subject we obtain

\mu=\frac {mg}{N} and substituting m for 38 Kg, g for 9.81 m/s^{2} and 170 N for  F we obtain

\mu=\frac{170} {38*9.81}=0.456033049

Therefore, the coefficient of static friction is 0.456033049

5 0
2 years ago
In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

Explanation:

We have given time period of the pendulum when length is L_1 is T_1=0.95sec

And when length is L_2 time period T_2=1sec

We know that time period is given by

T=2\pi \sqrt{\frac{L}{g}}

So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

And 1=2\pi \sqrt{\frac{L_2}{g}}-------eqn 2

Dividing eqn 2 by eqn 1

\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}

Squaring both side

\frac{L_2}{L_1}=1.108

8 0
2 years ago
High‑speed ultracentrifuges are useful devices to sediment materials quickly or to separate materials. An ultracentrifuge spins
pantera1 [17]

Answer:

Fc = 7.14N

Explanation:

First of all, let's convert everything to the same unit system:

m = 0.0031kg     R = 13.1cm * 1m / 100cm = 0.131m      

ω = 50000 rev/min * 1rev /( 2π rad ) * 1min / 60s = 132.63 rad/s

Now we can calculate centripetal force as:

Fc = m * \frac{V^2}{R} = m * \frac{(\omega*R)^2}{R}=m*R*\omega ^2

Replacing the values we get the answer:

Fc = 7.14N

3 0
2 years ago
3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the
bulgar [2K]

Answer:

The wife have to sit at 0.46 L from the middle point of the seesaw.

Explanation:

We need to make a sketch of the seesaw and the loads acting over it.

And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.

A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.

The maximum momentum with respect to the pivot point (0) will be:

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Where L/2 is the half of the distance of the seesaw

Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.

7 0
2 years ago
A 10.0 cm3 sample of copper has a mass of 89.6
Romashka-Z-Leto [24]
Density is mass divides by volume, so
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*cm^3 is a standard unit of volume*
4 0
2 years ago
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