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zvonat [6]
3 years ago
11

What do we call the force used when someone is pushing a lawnmower across the yard?

Physics
2 answers:
raketka [301]3 years ago
7 0
C.) applied force is the answer

dimaraw [331]3 years ago
7 0
I think your answer is C. applied, that is the force we use when pushing a lawnmower across the yard. 
Hope this helps.
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In which situation does Newton’s third law play a role?
yanalaym [24]

Answer: B

Explanation:

7 0
3 years ago
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Sarah is hanging a 1.5 -kg picture on the wall. She holds it in place with a horizontal force while she checks to see if it is p
NNADVOKAT [17]

Answer:

24.5 N

Explanation:

Draw a free body diagram of the picture.  There are four forces:

Weight force mg pulling down.

Friction force Nμ pushing up.

Normal force N pushing left.

Applied force F pushing right.

Sum of the forces in the y direction:

∑F = ma

Nμ − mg = 0

N = mg / μ

Sum of the forces in the x direction:

∑F = ma

F − N = 0

F = N

Substitute and solve:

F = mg / μ

F = (1.5 kg) (9.8 m/s²) / (0.6)

F = 24.5 N

4 0
3 years ago
With a wooden ruler, you measure the length of a rectangular piece of sheet metal to be 14 mm. With micrometer calipers, you mea
ipn [44]

Answer:

Follows are the solution to the given question:

Explanation:

The rectangular part has a length of 14 \ mm and its rectangular part has a width of 4.98 \ mm.

In option A

Calculating the area of the rectangular throgh the given piece:

\to A_R = WL=(14 mm) (4.98 mm) =69.72 \ mm^2

In option B

Calculating the ratio of rectangle's width which is rectangle's length:

\to R_{WL}=\frac{W}{L}= \frac{4.98 \ mm}{14 \ mm} = 0.3557

So, the ratio of rectangle's width to rectangle's length is 0.3557  .

In option C

Calculating the Perimeter of the rectangle:

\to  P_R=2(W+L)=2(14 \ mm+ 4.98 \ mm)= 2(18.98) = 37.96 \ mm  

In option D

Calculating the difference between length and width:  

\to D_{LW} = L- W = 14\  mm -4.98 \ mm =9.02 \ mm

In option E

Calculating the ratio of length to width:

\to R_{LW}=\frac{L}{W} =\frac{14\ mm}{4.98 \ mm} = 2.811

4 0
3 years ago
(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.4 mm in diameter in a steel alloy when a load of 1000
ruslelena [56]

Answer:

(a). The HB of this material is 217.8

(b). The diameter of an indentation is 1.45 mm.

Explanation:

Given that,

Diameter of Brinell hardness D= 10.0 m

Diameter of steel alloy d= 2.4 mm

Load = 1000 kg

(a). We need to calculate the HB of this material

Using formula of Brinell hardness

HB=\dfrac{2P}{\pi D(D-\sqrt{D^2-d^2})}

Put the value into the formula

HB=\dfrac{2\times1000}{\pi\times10(10-\sqrt{10^2-2.4^2})}

HB=217.8

(b). Given that,

Hardness = 300 HB

Load = 500 kg

We need to calculate the diameter of an indentation

Using formula of diameter

d=\sqrt{D^2-[D-\dfrac{2P}{(HB)\pi D}]^2}

Put the value into the formula

d=\sqrt{10^2-[10-\dfrac{2\times500}{300\pi\times10}]^2}

d=1.45\ mm

Hence,(a). The HB of this material is 217.8

(b). The diameter of an indentation is 1.45 mm.

7 0
3 years ago
The table lists the mass and charge of a proton and a neutron.
Olegator [25]

The gravitational force is much larger than the electrical force for any distance between the particles.

Explanation:

The gravitational force between two particles is given by:

F=\frac{Gm_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the two particles

r is the separation between the two particles

The electric force between two particles is given by:

F=\frac{kq_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the two particles

In this problem, we are comparing the gravitational force and the electric force between a proton and a neutron.

We must immediately notice that the charge of the neutron is zero, since it is electrically neutral:

q = 0

This means that the electric force between the proton and the neutron is always zero, regardless of their distance. And therefore, since their masses is not zero (and the gravitational force is never zero), the correct answer is

The gravitational force is much larger than the electrical force for any distance between the particles.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

About electrical force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
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