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ozzi
3 years ago
14

a rod of zinc has a mass of 5kg. in water it weighs 25 N. what is the buoyant force on it when it is submerged in water?

Physics
1 answer:
Alexxandr [17]3 years ago
7 0

Answer:

If it has a mass of 5 kg its weight is 5 kg * 9.8 m/s^2 = 49 N

Weight in air = 49 N

Weight in water = 25 N

Buoyant force provided by water = 49 - 25 = 24 N

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A 40kg mass is pulled by a horizontal force of 200N, with a constant velocity. What is h^2?
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Answer:

40,000

Explanation:

200x 200=40000

I'm pretty sure that's the answer

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Security is a major concern now that computers play such a large part in healthcare. To address this concern, as well as others
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The correct answer is C. HIPAA

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What mass of a material with density ρ is required to make a hollow spherical shell having inner radius r1 and outer radius r2?
krok68 [10]

Answer: m=\frac {\rho}{\frac{4}{3} \pi( (r_2)^{3} - (r_1)^{3}) }

It is given that:

Density of material =ρ

radius of inner spherical shell = r_1

radius of outer spherical shell=r_2

we know that the volume of sphere is \frac{4}{3}\pi r^{3}

Volume of the given spherical shell = \frac{4}{3}\pi( (r_2)^{3}-(r_1)^{3})

Then mass of the spherical shell can be calculate as:

Mass, m=density/ volume

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6 0
3 years ago
A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 73° above the hor
Brrunno [24]

Answer:

7 s

Explanation:

u = Initial velocity of rock = 12.2 m/s

\theta = Angle of throw = 73^{\circ}

x = Displacement in x direction = 25 m

Displacement in x direction is given by

x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{25}{12.2\times \cos73^{\circ}}\\\Rightarrow t=7\ \text{s}

Time taken to reach the ground is 7 s.

4 0
3 years ago
A concrete highway is built of slabs 12 m long (20°C). How wide should the expansion cracks between the slabs be (at 20°C) to pr
Bezzdna [24]

Answer:

0.001152m

Explanation:

Linear expansivity of a material is the change in length of the material per unit length per degree rise in temperature. Mathematically,

¢ = ∆L/L1∆°C

¢ is the linear expansivity of the material = 12 x 10⁻⁶ °C⁻¹

Where ∆L is the change in length = L2-L1

L2 is the final length = ?

L1 is the initial length = 12m

∆°C is the change in temperature = °C2 - °C1 = 50-(-30) = 80°C

Substituting this values inside the formula to get the final length L2 after expansion, we have;

12 x 10⁻⁶ °C⁻¹ = L2-12/12×80

12 x 10⁻⁶ °C⁻¹ = L2-12/960

L2-12= 960×12 x 10⁻⁶ °C⁻¹

L2-12 = 0.001152

L2 = 12+0.001152

L2 = 12.001152m

Expansion will be the change in length L2-L1 = 12.001152-12

= 0.001152m

The expansion cracks between the slabs should be 0.001152m wide to prevent buckling

5 0
3 years ago
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