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3 years ago

PLEASE HELP I WILL GUVE BRAINLY

Physics

1 answer:

katen-ka-za [31]3 years ago

8 0

Answer:

While they might look intimidating, the battle ropes are a simple workout tool that everyone can use. You'll train the muscles in your upper back, arms, abs, back, glutes and if you incorporate jumps, lunges, and squats, you can also work your legs!

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A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

$T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m$

Hence the force constant of the spring is 17.71N/m

4 0

3 years ago

Bob is pulling a 30 kg filing cabinet with a force of 200 N , but the filing cabinet refuses to move. The coefficient of static

Arada [10]

Answer:

<em>Magnitude of the Frictional force is 200 N</em>

Explanation:

The frictional force is the force that tries to oppose relative motion between two surfaces that are contacting. The coefficient of static friction is the coefficient of friction of a body that is not moving.

Newton's third law of motion states that action and reaction forces are equal and opposite. So the frictional force felt on the filing cabinet will be equal to the applied force pulling the cabinet.

Frictional force = Force applied

Force applied = 200 N

Therefore, the magnitude of the friction force on the filing cabinet is 200 N

5 0

4 years ago

A starship travels to a planet that is 20 light years away. The astronauts stay on the planet for 2.0 years before returning at

ad-work [718]

Answer:

astronauts age is 32 years

correct option is e 32 years

Explanation:

given data

travels = 20 light year

stay = 2 year

return = 52 years

to find out

astronauts aged

solution

we know here they stay 2 year so time taken in traveling is

time in traveling = ( 52 -2 ) = 50 year

so it mean 25 year in going and 25 years in return

and distance is given 20 light year

so speed will be

speed = distance / time

speed = 20 / 25 = 0.8 light year

so time is

time = $\frac{t}{\sqrt{1-v^2} }$

time = $\frac{25}{\sqrt{1-0.8^2} }$

time = 15 year

so age is 15 + 2 + 15

so astronauts age is 32 years

so correct option is e 32 years

4 0

3 years ago

If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c

Lemur [1.5K]

Answer:

$\mu_s \geq 0.27$

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

$f_s=\frac{mv^{2}}{R}$

But we know that:

$f_s\leq \mu_s N$

And the normal force is given by the sum of the forces in the vertical direction:

$N-mg=0 \implies N=mg$

Finally, we have:

$f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27$