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artcher [175]
3 years ago
6

If the mass of a material is 50 grams and the volume of the material is 12 cm^3, what would the density of the material be?

Physics
1 answer:
AlekseyPX3 years ago
4 0
The density of the material would be 
25/6 grams per cm^3.
to obtain the result above this is what we do:
density is calculated as: (the mass of the given material or object) / volume of the material 
which leads us to 50grams /12cm^3
 
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An arrow movirg 48.3 m/s has 5.22<br> kg•m/s of momentum. What is its<br> mass?
spin [16.1K]

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

6 0
3 years ago
Two dogs fight over a bone. The larger of the two pulls on the bone to the right with a force of 42 N. The smaller one pulls to
tamaranim1 [39]

Answer:

Explanation:

A

35 N  Small Dog <=======BONE=========> Bigger Dog 42 N

B

Fnet = Large Dog - small dog The forces are subtracted because they are acting in opposite Directions.

Fnet = 42 - 35

Fnet = 7 N

C

m = 2.5 kg

F = 7 N

a = ?

F = m * a

7 = 2.5 a

a = 7 / 2.5

a = 2.8 m/s^2

4 0
2 years ago
The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee
lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is -27.3\hat{i}

y component of vector is 43.6\hat{j}

so position vector is

r=-27.3\hat{i}+43.6\hat{j}

Magnitude of vector is

|r|=\sqrt{27.3^2+43.6^2}

|r|=\sqrt{2646.25}

|r|=51.44 units

Direction

tan\theta =\frac{43.6}{-27.3}=-1.597

vector is in 2nd quadrant thus

180-\theta =57.94

\theta =122.06^{\circ}

4 0
3 years ago
At which temperature does the motion of atoms and molecules stop?<br> 0°C<br> 0C<br> 0°K<br> 0K
anastassius [24]

Answer: 0K

Explanation:

Absolute 0 (0K) is the point where nothing could be colder and no heat energy remains in a substance.

7 0
3 years ago
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the ri
lawyer [7]

Answer:

x = 0.6034 m

Explanation:

Given

L = 5 m

Wplank = 225 N

Wman = 522 N

d = 1.1 m

x = ?

We have to take sum of torques about the right support point.  If the board is just about to tip, the normal force from the left support will be going to zero.  So the only torques come from the weight of the plank and the weight of the man.

∑τ = 0  ⇒     τ₁ + τ₂ = 0  

Torque come from the weight of the plank = τ₁

Torque come from the weight of the man = τ₂

⇒  τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)

⇒  τ₂ = Wman*x = 522 N*x   (clockwise)

then

τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0

⇒  x = 0.6034 m

7 0
3 years ago
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