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2 years ago

What is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 30 m/sec

1 answer:

5 0

To find this we should use the equation Ek=1/2mv^2 where Ek is the kinetic energy, m is the mass and v is the velocity. So by this Ek=1/2*1*30^2 which then makes Ek=450 joules

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Why do most scientists follow a set order of steps when carrying out a scientific investigation?

eimsori [14]

Scientists follow a set order of steps when carrying out a scientific investigation to make sure that the method, interpretation and results that they have obtained are repeatable and reliable. This kind of information can be truly said that their data is true and valid.

6 0

2 years ago

Read 2 more answers

Consider a sphere and a cylinder of equal volume made of copper. Both the sphere and the cylinder are initially at the same temp

zepelin [54]

Answer: The cylinder

Explanation:

Among all other solid shapes, the sphere has the smallest area for a given volume.

By experiment, the ratio of the radius of a sphere to a cylinder of equal volume is less than 1.

Recall;

That the Rate of transfer of convective heat (Q) = h × A ×change in temperature.

Where ,

h= the co efficient of convective heat transfer

A= the cross sectional area.

As such, since the sphere has a smaller surface area relative to the cylinder, the sphere transfers heat slower than the cylinder.

Therefore, if the sphere and cylinder are exposed to convection in the same environment, then, the cylinder cools faster.

PS; the more the Area, the higher the rate of heat transfer and vice versa.

7 0

2 years ago

The x-component of vector R is Rx = −28.2 units and its y-component is Ry = 19.6 units. What are its magnitude and direction? Gi

ikadub [295]

Answer:

Explanation:

Rx = -28.2 units

Ry = 19.6 units

magnitude of R = √ [( - 28.2 )² + ( 19.6 ) ]

= √ ( 795.24 + 384.16 )

= 34.34 units

If θ be the angle measured counterclockwise from the +x-direction

Tanθ = 19.6 / - 28.2 = -0.695

θ = 180 - 34.8

= 145.2° .

7 0

3 years ago

A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it

joja [24]

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

$U = F.s$

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

$30 = F(1.5)$

$F = 20 N$

6 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of 4.15 m from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

3 0

2 years ago