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3 years ago

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A horizontal force of 675 N is needed to overcome the force of static friction between a level floor and a 300-kg crate. What is

the acceleration of the crate if the 675-N force is maintained after the crate begins to move and the coefficient of kinetic friction is 0.15

1 answer:

White raven [17]3 years ago

8 0

Explanation:

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If a freely falling rock were equipped with a speedometer, by how much would its speed readings increase with each second of fal

Klio2033 [76]

Its speed reading would increase to 10 m/s every second

5 0

3 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

igomit [66]

Answer:

The maximum electric power output is $P_{max} =1.339*10^{9} \ W$

Explanation:

From the question we are told that

The capacity of the hydroelectric plant is $\frac{V}{t} = 690 \ m^3 /s$

The level at which water is been released is $h = 220 \ m$

The efficiency is $\eta =$ 0.90

The electric power output is mathematically represented as

$P = \frac{PE_l - PE _o}{t}$

Where $PE_l$ is the potential energy at level h which is mathematically evaluated as

$PE_l = mgh$

and $PE_o$ is the potential energy at ground level which is mathematically evaluated as

$PE_o = mg(0)$

$PE_o = 0$

So

$P = \frac{mgh}{t}$

here $m = V * \rho$

where V is volume and $\rho$ is density of water whose value is $\rho = 1000 kg/m^3$

So

$P = \frac{(\rho * V) * gh}{t}$

$P = \frac{V}{t} * gh \rho$

substituting values

$P =690 * 9.8 * 220 * 1000$

$P =1.488*10^{9} \ W$

The maximum possible electric power output is

$P_{max} = P * \eta$

substituting values

$P_{max} =1.488*10^{9} * 0.90$

$P_{max} =1.339*10^{9} \ W$

6 0

3 years ago

While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would

Pavel [41]

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

$F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N$

$\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\$

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

$\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\$

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

4 0

3 years ago

The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?

Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours. So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes. So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0

3 years ago

(4.3 x 10-2 )4.950 x 105)<br> O 21.285 x 103<br> O 2.13 x 104<br> O 21 x 103<br> O 2.1 x 104

Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

7 0

3 years ago