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sergeinik [125]
3 years ago
12

Use the following terms in the same sentence:proton, neutron ,and isotope

Physics
1 answer:
UkoKoshka [18]3 years ago
8 0
Atoms of the same element having equal numbers of protons, but different numbers of neutrons is called isotope.
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Answer <br> A<br> B<br> C<br> D and <br> E
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A line with a negative slope on a distance-time graph means the object is?​
otez555 [7]

Answer:negative slope or gradient I think

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Read 2 more answers
Calculate the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of
vesna_86 [32]

Answer:

0.21%

Explanation:

We are given;

Mass; m = 100 kg

Diameter; d = 2.2 mm = 2.2 × 10^(-3) m

Young's modulus; E = 12.5 x 10^(10) N/m².

Formula for area is;

A = πd²/4

A = (π/4) x (2.2 x 10^(-3))²

A = 3.8 x 10^(-6) m²

Force; F = mg

g is acceleration due to gravity and has a constant value of 9.8 m/s²

F = 100 × 9.8

F = 980 N

Formula for young's modulus is;

E = Stress/strain

Formula for stress = F/A

Formula for strain = ΔL/L

Thus;

E = (F/A)/(ΔL/L)

Making ΔL/L the subject, we have;

ΔL/L = (F/A)/E

Plugging in the relevant values;

ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))

ΔL/L = 0.0021

Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%

8 0
3 years ago
A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00
Oxana [17]

Answer:

12.97 km

Explanation:

In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.

Taking north as positive y direction and east as positive x-direction, we have:

- Displacement 1: 2.00 km to the north

So

A_x = 0\\A_y = +2.00 km

- Displacement 2: 60.0° south of east for 7.00 km

So

B_x=(7.00)(cos (-45^{\circ}))=4.95 km\\B_y = (7.00)(sin (-45^{\circ}))=-4.95 km

- Displacement 3: 9.50 km 35.0° north of east

So

C_x=(9.50)(cos 35^{\circ})=7.78 km\\C_y=(9.50)(sin 35^{\circ})=5.45 km

So the net displacement along the two directions is:

R_x=A_x+B_x+C_x=0+4.95+7.78=12.73 km\\R_y=A_y+B_y+C_y=2.00+(-4.95)+5.45=2.50 km

So, the  distance between the initial and final position is equal to the magnitude of the net displacement:

R=\sqrt{R_x^2+R_y^2}=\sqrt{12.73^2+2.50^2}=12.97 km

7 0
3 years ago
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