An object is moving to the left and is slowing down. its acceleration is

Assume that segment r exerts a force of magnitude t on segment l. what is the magnitude flr of the force exerted on segment r by

mrs_skeptik [129]

If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick $T_R=T$ whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that $T_L=-T$

8 0

3 years ago

a crane lifts a 75 kg mass at a height of 8m. calculate the gravitational potential energy gained by the mass (g= 10N)

Leno4ka [110]

Answer:

6000 J

Explanation:

Potential energy = mass*gravity*height (U = m*g*h)

In this case g = 10 so

U = 75*10*8 = 6000 J (joules) [kg * m^2/s^2]

7 0

3 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

4 years ago

. A newly discovered planet has three times the mass and five times the radius of Earth. What is the ratio of the acceleration d

NikAS [45]

Answer:

0.12

Explanation:

The acceleration due to gravity of a planet with mass M and radius R is given as:

g = (G*M) / R²

Where G is gravitational constant.

The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg

The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m

Therefore:

g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²

g(planet) = 1.18 m/s²

Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:

g(planet)/g(earth) = 1.18/9.8 = 0.12

3 0

3 years ago

Am I correct?? Will give brainliest

IRISSAK [1]

You are correct earth science is studied to predict planetery changes

4 0

3 years ago

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