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vazorg [7]
3 years ago
15

A block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. The block weighs 1587 g. From this information, calculate the d

ensity of lead.
Physics
1 answer:
mamaluj [8]3 years ago
6 0

Answer:

11.3 g/cm^3

Explanation:

density = mass/volume

volume of rectangular prism = length * width * height

volume = (4.50 cm)(5.20 cm)(6.00 cm) = 140.4 cm^3

mass = 1587 g

density = (1587 g)/(140.4 cm^3)

density = 11.3 g/cm^3

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Give two examples of direct current electricity.​
ludmilkaskok [199]

Answer:

<h3>Ion beams and Electrochemical are two examples of direct current electricity.</h3>
6 0
3 years ago
A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft
matrenka [14]

Answer:

The velocity of the players will be <u>2.88 m/s</u> in the <u>east</u> direction.

Explanation:

Let 'v' be the velocity of the players after collision.

Consider the east direction as positive direction.

Given:

Mass of the first player  is, m_1 = 91.5 kg  

Initial velocity of the first player  is, u_1 = 2.73 m/s  

Mass of the second player  is, m_2 = 63.5 kg  

Initial velocity of the second player is, u_2 = 3.09 m/s  

In order to solve this problem we use the law of conservation of momentum which says that the total momentum must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v

Solving for v, we get:

v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(2.73)+(63.5)(3.09)}{91.5+63.5}=2.88\ m/s

Therefore, their velocity after the collision is 2.88 m/s.

The sign of the velocity after collision is positive. So, the players will move in the east direction only after collision.

3 0
3 years ago
A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction betwe
Gre4nikov [31]

Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

            28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

5 0
3 years ago
What is the total amount of force needed to keep a 6.0 kg object moving at speed
DanielleElmas [232]

Answer:

C. 0 N

Explanation:

In the absence of external forces, a body in motion will stay in motion.

F = ma

F = 6.0(0) = 0

8 0
3 years ago
Read 2 more answers
A car is accelerating at 30 m/s2, if the car is 400 kg how much force
Verizon [17]
It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .

please mark brainliest and i hope this helps!
6 0
3 years ago
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