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Salsk061 [2.6K]

3 years ago

14

A rock has a mass of 3.1 kg. What is its weight on earth

1 answer:

Masteriza [31]3 years ago

6 0

Answer:

W = 30.38 N

Explanation:

Given that,

Mass of a rock, m = 3.1 kg

We need to find the weight of the rock on the surface of Earth. Weight of an object is given by :

W = mg

g is the acceleration due to gravity, g = 9.8 m/s²

W = 3.1 kg × 9.8 m/s²

= 30.38 N

So, the weight of the rock on the Earth is 30.38 N.

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A mass attached to a spring oscillates in simple harmonic motion with an amplitude of 10 cm. When the mass is 5.0 cm from its eq

timama [110]

When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

<h3>Total energy of the mass</h3>

The total energy possessed by the mass under the simple harmonic motion is calculated as follows;

U = ¹/₂kA²

where;

k is the spring constant
A is the amplitude of the oscillation

<h3>Potential energy of the mass at 5 cm from equilibrium point</h3>

P.E = ¹/₂k(Δx)²

<h3>Kinetic energy of mass</h3>

K.E = U - P.E

K.E = ¹/₂kA² - ¹/₂k(Δx)²

<h3>Percentage of its energy that is kinetic</h3>

$K.E (\%) = \frac{U - P.E}{U} \times 100\%\\\\K.E (\%) =\frac{\frac{1}{2}kA^2 - \frac{1}{2}k(\Delta x)^2 }{\frac{1}{2}kA^2} \times 100\%\\\\K.E (\%) = \frac{A^2 - (\Delta x)^2}{A^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - (10-5)^2}{10^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - 5^2}{10^2} \times 100\%\\\\K.E (\%) = 75\%$

Thus, when the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

Learn more about kinetic energy here: brainly.com/question/25959744

3 0

2 years ago

A flat roof is very susceptible to wind damage during a thunderstorm and/or tornado. If a flat roof has an area of 500 m2 and wi

Evgesh-ka [11]

Answer: The magnitude of the force exerted on the roof is 490522.5 N.

Explanation:

The given data is as follows.

Below the roof, $v_{1}$ = 0 m/s

At top of the roof, $v_{2}$ = 39 m/s

We assume that $P_{1}$ is the pressure at lower surface of the roof and $P_{2}$ be the pressure at upper surface of the roof.

Now, according to Bernoulli's theorem,

$P_{1} + 0.5 \times \rho \times v^{2}_{1} = P_{2} \times 0.5 \rho \times v^{2}_{2}$

$P_{1} - P_{2} = 0.5 \times \rho \times (v^{2}_{2} - v^{2}_{1})$

= $0.5 \times 1.29 \times [(39)^{2} - (0)^{2}]$

= $0.645 \times 1521$

= 981.045 Pa

Formula for net upward force of air exerted on the roof is as follows.

F = $(P_{1} - P_{2})A$

= $981.045 \times 500$

= 490522.5 N

Therefore, we can conclude that the magnitude of the force exerted on the roof is 490522.5 N.

5 0

3 years ago

All ions are atoms with a?

konstantin123 [22]

I think it would be B

4 0

3 years ago

A uniform ladder of length l rests against a smooth, vertical wall. If the coefficient of static friction is 0.50, and the ladde

Ronch [10]

Answer:

X = 5.44 m

Explanation:

First we can calculate the normal force acting from the floor to the ladder.

W₁+W₂ = N

W1 is the weigh of the ladder

W2 is the weigh of the person

So we have:

$m1g+m2g=N$

$N=755.37 N$

The friction force is:

$F_{force}=\mu N=0.5\cdot 755.37=377.68 N$

Now let's define the conservation of torque about the foot of the ladder:

$\tau_{ledder}+\tau_{person=\tau_{reaction}}$

$m_{1}\cdot g\cdot X \cdot cos(53)+m_{2}\cdot g\cdot 3.75 \cdot cos(53)=F_{force}7.5sin(53)$

Solving this equation for X, we have:

$X = \frac{377.68\cdot 7.5\cdot sin(53)-21\cdot 9.81\cdot 3.75 \cdot cos(53)}{56\cdot 9.81\cdot cos(53)}$

Finally, X = 5.44 m

Hope it helps!

3 0

3 years ago

The staples inside a stapler are kept in place by a spring with a relaxed length of 0.116 m. if the spring constant is 46.0 n/m,

ankoles [38]

Answer:

U = 0.0207 J

Explanation:

We know that:

U = $\frac{1}{2}Kx^2$

where U is the potential energy, K the constant of the spring and x is the deformation.

so, the deformation is calcualted as:

x = 0.146m-0.116m

x = 0.03m

Finally, replacing the values of x and K, we get:

U = $\frac{1}{2}(46n/m)(0.03m)^2$

U = 0.0207 J

8 0

4 years ago