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BlackZzzverrR [31]
3 years ago
5

Which of the following are NATURAL resources? Which of the following are NATURAL resources? Select all that apply.

Physics
2 answers:
scoundrel [369]3 years ago
4 0
Wood, water, Forests. The rest can have harm to the nature.
docker41 [41]3 years ago
3 0

Minerals

Forests

Wood

Water

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5. What is the density of 4.5 mL of a liquid that has a mass of 1.3 grams?
antoniya [11.8K]

Answer:

A. 0.289g/mL

Explanation:

Using the equation for density which is d = m/v  or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.

5 0
3 years ago
Read 2 more answers
2. What do you understand by balanced and unbalanced force​
labwork [276]

Answer:

forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.

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hope helpful ~

8 0
2 years ago
A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
3 years ago
NEED ANSWER ASAP!!! Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket.
Schach [20]

a. The free-body diagram for the glass when it is at the top of the circle is attached below.

b. The equation for the net force on the glass at the top of the circle in terms of w, Fn, m, v, and r is mg x g + N -  mg x Vtop² /R =0

c. The glass will fall out of the bucket if the normal force between the glass and bucket equals zero. The speed with which she spin the bucket to prevent this from happening is 3.83 m/s.

d. The string will break if the tension on it is more than 100 N. The range of speeds can  prevent the string from breaking is 3.83< Vtop<4.99 m/s

<h3>What is Net force?</h3>

When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.

Given, Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket. She spins the bucket in a vertical circle of radius 1.5 m. She must swing the bucket to keep the glass from falling out.

a. The free body diagram of the bucket and glass is attached below.

b. Bucket will undergo centrifugal force

Fb = mVtop² /R

From the equilibrium of forces, we have

For bucket,

T +mb xg - N =  mb x Vtop² /R..............(1)

For glass,

mg x g + N =  mg x Vtop² /R..............(2)

Thus, this is the net force equation on the glass.

c. On adding both the equations. we have

T + (mb + mg) xg = (mb + mg) Vtop² /R

Substituting the values, T = 0 and from the question, we get

0 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)

Vtop = 3.83 m/s

Thus, the speed of spin to prevent glass from falling out is 3.83 m/s

d. The string will break if the tension on it is more than 100 N

100 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)

Vtop = 4.99 m/s

Thus, the range of velocity is  3.83< Vtop<4.99 m/s

Learn more about net force.

brainly.com/question/18031889

#SPJ1

8 0
2 years ago
A football is thrown horizontally with an initial velocity of(16.6 {\rm m/s} ){\hat x}. Ignoring air resistance, the average acc
Ray Of Light [21]

Answer:

A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s  c) 46º below horizontal.

Explanation:

A) Once released, the football is not under the influence of any external force in the horizontal direction, so it  continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.

If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that  this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:

vₓ = 16.6 m/s i

In the vertical direction, the football, once released, is in free fall, starting from rest.

So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:

vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s

Assuming that the upward direction is the positive  for the y-axis (perpendicular to the chosen  x-axis), we can write the vertical component of  the velocity vector, at t=1.75 s, as follows:

vy = -17.2 m/s j

So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:

v = 16.6 m/s i -17.2 m/s j

b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:

v =\sqrt{(16.6m/s)^{2}+ (-17.2m/s)^{2}} = 23.9 m/s

v = 23.9 m/s

c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:

tg θ =\frac{-17.2}{16.6} =-1.036

⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.

6 0
3 years ago
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