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3 years ago

Name 2 forces acting on an object at rest Only need 1 got gravity

1 answer:

mote1985 [20]3 years ago

7 0

You mean like a box sitting on a table.

One force is the force of gravity, pulling downward on the box.

Now, you know that the forces acting on the box must be balanced, because
if they're not, then the box would be accelerating. But it's just sitting there, so
there must be some other force, just exactly the right strength and direction to
exactly cancel the force of gravity on the box, so that the net force on it is zero.

The other force is the force of the table pushing upward on the box. It's called
the "normal force".

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A 0.76-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one

sergey [27]

Answer:

1.54 kg

Explanation:

mass of first block (m) = 0.76 kg

acceleration due to gravity (g) = 9.8 m/s

what is the mass (m) of the second block

force = kx ⇒ mg = kx

mg = kx

where m is the mass, g is the acceleration due to gravity, k is the

spring constant and x is the extension

0.76 x 9.8 = kx

7.5 = kx

k = 7.5/x ... equation 1

when a second block is attached to the first one the amount of stretch triples (this means that extension (x) = 3x)

therefore the new mass becomes m + 0.76 and the extension

becomes 3x

with the new mass and extension, mg = kx now becomes

(m+0.76)g = k(3x) ... equation 2

Recall that k = 7.5/x from equation 1, substituting this value of k into

equation 2 we have

(m+0.76)g = $\frac{7.5}{x}$ × (3x)

(m+0.76)g = 7.5 × 3

substituting the value of g = 9.8 m/s^{2}

(m + 0.76) x 9.8 = 7.5 x 3

m + 0.76 = 22.5 ÷ 9.8

m + 0.76 = 2.3

m = 2.3 - 0.76 = 1.54 kg

4 0

3 years ago

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

konstantin123 [22]

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

$L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}$

Where

$I_{i}\ and \ \omega_{i}$ initial moment of inertia and angular velocity

$I_{f}\ and \ \omega_{f}$ is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ is the satellite mass

r is the radus of the sphere

Substititute 1900kg for m and 4.6m for r

$I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}$

The final moment of inertia of the satellite about the centre of mass

$I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}$

Where $m_{x}$ is the antenna's mass and

I is the length of the antenna

$I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}$

So, the Final rotation rate of the satellite is:

$I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

5 0

3 years ago

If the total mass is m, find the moment of inertia about an axis through the center and perpendicular to the plane of the square

My name is Ann [436]

Question: A thin, uniform rod is bent into a square of... A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem.

5 0

3 years ago

*WILL MARK BRAINLIEST*

Arte-miy333 [17]

Answer:

1.52

Explanation:

5 0

3 years ago

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The specific heat of Aluminum is 0.9 J/g K. The specific heat of Copper is 0.39 J/g K. If samples of equal mass of both Aluminum

AleksandrR [38]

Answer:The Aluminum loses a little more than twice the heat of the Copper.Explanation:
Since specific heat is part of the equation. A smaller specific heat will create a smaller heat gain or loss. 
Hope this helped!!!!

6 0

3 years ago

Read 2 more answers