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yarga [219]
3 years ago
9

Three boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass i

n the center. A force of 50 N pushes on the 5.0-kg box, which pushes against the other two boxes. What magnitude force does the 5.0-kg box exert on the 3.0-kg box?

Physics
1 answer:
Fed [463]3 years ago
6 0

Answer:25 N

Explanation:

Given

mass of 5 , 3 & 2 kg blocks lie on floor

Force on 50 N pushes the 5 kg box

Let N_1 be the reaction on 2 kg box therefore

N_1=2\times a

where a is the acceleration of the system

N_2=reaction on 3 kg block by 5 kg block

N_2-N_1=3a

N_2=5a

Now for 5 kg block

F-N_2=5a

F=10a

a=\frac{F}{10}=5 m/s^2

Force exerted by 5 kg block on 3 kg block is

N_2=5\times 5=25 N

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A student measured the mass of ice in a glass container with a tight lid. He allowed the ice to melt, and then found the mass of
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Answer:

The liquid formed from a melted solid has the same mass as the solid has.

Explanation:

As long as no water can escape, the mass of the ice before melting must equal the mass of the liquid water after.

7 0
3 years ago
Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Thr
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Answer:

7.2N/C

Explanation:

Pls see attached file

6 0
3 years ago
The free-body diagram of a crate is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labe
satela [25.4K]

The net force acting on the crate is determined as 176 N to the left.

<h3>Net force acting on the crate</h3>

The net force acting on the crate is calculated as follows;

∑F = F1 + F2 + F3 + F4

F(net) = -440y + 176x + 440y - 352x

F(net) = -176 x

The resultant force is pointing in negative x direction.

Thus, the net force acting on the crate is determined as 176 N to the left.

Learn more about net force here: brainly.com/question/14361879

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8 0
2 years ago
A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0
3 years ago
Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her h
mojhsa [17]

Answer:

Isabella will not be able to spray Ferdinand.

Explanation:

We'll begin by calculating the time taken for the water to get to the ground from the hose held at 1 m above the ground. This can be obtained as follow:

Height (h) = 1 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =.?

h = ½gt²

1 = ½ × 9.8 × t²

1 = 4.9 × t²

Divide both side by 4.9

t² = 1/4.9

Take the square root of both side

t = √(1/4.9)

t = 0.45 s

Next, we shall determine the horizontal distance travelled by the water. This can be obtained as follow:

Horizontal velocity (u) = 3.5 m/s

Time (t) = 0.45 s

Horizontal distance (s) =?

s = ut

s = 3.5 × 0.45

s = 1.58 m

Finally, we shall compare the distance travelled by the water and the position to which Ferdinand is located to see if they are the same or not. This is illustrated below:

Ferdinand's position = 10 m

Distance travelled by the water = 1.58 m

From the above, we can see that the position of the water (i.e 1.58 m) and that of Ferdinand (i.e 10 m) are not the same. Thus, Isabella will not be able to spray Ferdinand.

8 0
3 years ago
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