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lys-0071 [83]
2 years ago
5

at the sewing store, ava bought a bag of buttons. 21 in all. 21 of the buttons were large. what percentage of the buttons wre la

rge
Physics
2 answers:
erma4kov [3.2K]2 years ago
6 0
If she only has 21 buttons and all 21 of them are large, then all of her buttons are large. so 100% of the buttons would be large.
lukranit [14]2 years ago
4 0

Answer: 100%

Explanation:

Given : The total number of buttons in the bag= 21

The number of larger buttons = 21

Then , the percentage of the buttons were large is given by :-

\dfrac{\text{Number of large buttons}}{\text{Total buttons}}\times100\\\\\dfrac{21}{21}\times100=100\%

Therefore, the percentage of the buttons were large = 100%

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Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0
frez [133]
Since his line of sight 63 degrees makes with the tip of the building

Tan63° =  height of building / Horizontal distance

tan63° =  Height / 50

50tan63° = Height

Height = 50tan63°

Height ≈ 50*1.9626

Height ≈ 98.13 m

Height of the building is ≈ 98.13 m. Mind you in solving for this height we have neglected the height of Daniel.

The height of building actually should be 98.13 m plus the height of Daniel. Since the 63° was measured from his eye level.
4 0
2 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
2 years ago
A mover hoists a 50 kg box from the ground to a height of 2 m. What was the change in the box's energy
SSSSS [86.1K]

Answer:

980 J

Explanation:

The change in box's energy is equal to its change in gravitational potential energy:

\Delta U = m g \Delta h

where

m = 50 kg is the mass of the box

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta h= 2m is the change in height of the box

Substituting numbers, we find

\Delta U = (50 kg)(9.8 m/s^2)(2 m)=980 J

3 0
2 years ago
Question 8 (4 points)
katrin2010 [14]

Answer:

The answer to your question is:

Explanation:

Data

Duane                                          Albert

d = 5 m ;  v = 3 m/s                     v = 4.2 m/s

a)                                                b)

Duane's                                      Albert's

d = 5 + (3)t                                  d = 4.2t

d = 5 + 3t

c)                            5 + 3t = 4.2t

                              4.2t - 3t = 5

                                      1.2t = 5

                                           t = 4.17 s

d)

Duane's

d= 5 + 3(4.17)

d = 17.51 m

Alberts

d = 4.2(4.17)

d = 17.51 m

4 0
2 years ago
Read 2 more answers
An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a
blsea [12.9K]

Answer: apparent weighlessness.


Explanation:


1) Balance of forces on a person falling:


i) To answer this question we will deal with the assumption of non-drag force (abscence of air).


ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).


iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).


2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.


3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.


Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

3 0
2 years ago
Read 2 more answers
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