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Paha777 [63]
3 years ago
15

Help me please !!!!!

Physics
2 answers:
bezimeni [28]3 years ago
8 0

Answer:

confounding cause they had exposure to many programmes

kipiarov [429]3 years ago
3 0

Answer:

Confounding should be your answer please mark brainliest

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A big kid pushes a little kid and the little kid falls backward. The big kid remains relatively stationary. What best explains t
postnew [5]
The big kid is the strongest ?
7 0
2 years ago
Read 2 more answers
3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What
WITCHER [35]

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

F_{net}=ma

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction F_f, opposing the motion of the box, so to the left

So we can write the net force as

F_{net}=F-F_f

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

a=0

Therefore

F_{net}=0

WHich means:

F-F_f=0

And therefore,

F_f=F=25 N

which means that the force of friction is also 25 N.

6 0
3 years ago
A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to
kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d  

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 =  192 + 56 d + 4d^2 - 192

93 =  56 d + 4d^2

4d^2 + 56 d - 93 = 0

d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}

d=\frac{-56\pm 87.72}{8}\

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0
3 years ago
How much heat is released when a 10.0-g sample of iron cools from 75.0°c to 25.5 °c? the specific heat capacity of iron is 0.449
GalinKa [24]
M = 10.0 g, the mass of the iron sample
ΔT = 75 - 25.2 = 49.5°C, the decrease in temperature
c = 0.449 J/(g-°C), the specific heat of iron

The heat released is
Q = m*c*ΔT
    = (10.0 g)*(0.449 J/(g-°C))*(49.5 C)
    = 222.255 J

Answer: 222.3 J (nearest tenth)

8 0
3 years ago
Explain what characterises a sound as noise or music​
natta225 [31]

Explanation:

Noise is sound with a continuous structure. Music is composed of sounds with a fundamental frequency and overtones. Noise is composed of sounds with frequencies that range continuously in value from as low as you can hear to as high as you can hear — not necessarily at equal intensity, however.

7 0
3 years ago
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