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1 year ago

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A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,

what is the ball’s weight on the surface of Planet X?

1 answer:

Trava [24]1 year ago

3 0

Given data:

* The mass of the ball is 2 kg.

* The gravitational field strength at the surface of planet X is 5 N/kg.

Solution:

The weight of the ball on the planet X is,

$W=ma$

where m is the mass of ball, a is the gravitational field strength,

Substituting the known values,

$\begin{gathered} W=2\times5 \\ W=10\text{ N} \end{gathered}$

Thus, the weight of the ball on the surface of planet X is 10 N.

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Complete the equation???

soldi70 [24.7K]

Here in nuclear reaction we can say that sum of neutrons and protons in reactant side and product side will be same always

Here mass number on the product side is given to us

so sum of mass number is given as

$A_1 + A_2 = 265 + 1 = 266$

now on the reactant side also the number must be same

$A_1' + A_2' = 58 + x$

now we will have

$58 + x = 266$

$x = 208$

Now number of protons on product side is given as

$P_1 + P_2 = 108 + 0$

now we also know that atomic number of Fe is 26

so now we will have

$P_1' + P_2' = 108$

$26 + P_2' = 108$

$P_2' = 82$

now the equation is given as

$_{26}^{58}Fe + _{82}^{208}Pb = _{108}^{266}Hs + _0^1X$

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2 years ago

A 58 kg skier is going down a 35 degree slope. The areaof each

maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

$N = mgcos\theta$

Since the skier is standing on two skis, his weight will be divide by two

$N' = \frac{mgcos\theta}{2}$

Pressure is given as the force applied in a given area, that is

$P = \frac{F}{A}$

Replacing F with N'

$P = \frac{N'}{A}$

$P = \frac{\frac{mgcos\theta}{2}}{A}$

Our values are given as,

$m = 58kg$

$g = 9.8m/s^2$

$\theta = 35\°$

$A = 0.3m^2$

Replacing we have that

$P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}$

$P = 776.01Pa$

Therefore the pressure exerted by each ski on the snow is 776.01Pa

6 0

3 years ago

In a tug of war, side A applies 10N force and side B applies 8N .which side will the rope move

choli [55]

The side B will fall down so that means the rope will move the direction of A

7 0

3 years ago

Steel is very stiff, and the Young's modulus for steel is unusually large, 2.0×1011 N/m2. A cube of steel 25 cm on a side suppor

TEA [102]

Answer:

Force (normal) = 833.85 N Compression = 1.67 x 10⁻⁸ m

Explanation:

Given data Young's Modulus (Y) = 2 x 10¹¹ N/m², Length of one side of cube = 25 cm = 0.25 m, mass of load = 85 kg

Normal force is the force exerted upon an object that is in contact with another stable object. This force would be applied by the surface onto the object in the same vector and is used to keep the object stable while it rests on a surface.

We know from Newton's Second Law that

F = ma where m is the mass and a is the acceleration (<em>in this case due to gravity</em>) hence, the normal opposing force to the load applied by the surface would be equal to the force applied on the surface by the weight of the load on the surface, So

F (normal) = M (load) x a = 85 x 9.81 = 833.85 N

Compression is the change in length of an object by the exertion of force upon it. Using the Young's Modulus formula we can find this change in the cube of steel. The Young's Modulus is given by

Y = (F/A)/(ΔL/L), where Y is the Young's Modulus, F is the Force being applied on the object, A is the cross sectional area on which the said force is applied, ΔL is the change in length due to said force being applied and L is the original Length of the side of the cross sectional area.

Solving this for ΔL, we can re- arrange the equation

ΔL = (F x L)/(Y x A) since area of square is L x L we can simplify the equation to get

ΔL = (F)/(Y x L), substitute the values

ΔL = (833.85)/(2 x 10¹¹ x 0.25) = 1.67 x 10⁻⁸m

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3 years ago

Which of the following absorbs the MOST heat energy?

goldfiish [28.3K]

Darker is warmer C. Black objects.

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2 years ago

Read 2 more answers