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2 years ago

Can someone please help me omg

Chemistry

2 answers:

saveliy_v [14]2 years ago

4 0

They are both heterozygous. If you look at the parents, each parent has one ‘A’ allele and one ‘a’ allele. When they have both alleles, they are heterozygous

Paladinen [302]2 years ago

3 0

I think option b is the answer

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4 years ago

In the Atomium building, what do the tubes represent?

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3 years ago

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3 years ago

Table salt is also known as sodium chloride, because it is

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1 year ago

A particle with a charge of 9.40 nC is in a uniform electric field directed to the left. Another force, in addition to the elect

juin [17]

Answer:

a. Work done by the electric force = -2.85 * ×10⁻⁵ J

b. The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. The magnitude of the electric field is 33.7kV/m

Explanation:

Given.

Charge = Q = 9.40 nC

Distance = d = 9.00 cm = 0.09m

Amount of work = 7.10×10⁻⁵ J

Kinetic energy = K = 4.25×10⁻⁵ J

a. What work was done by the electric force?

This is calculated by; change in Kinetic Energy i.e. ∆KE

∆KE = ∆K2 - ∆Kæ

Where K2 = 4.25×10⁻⁵ J

The body is released at rest, so the initial velocity is 0.

So, K1 = 0

Also, total work done = W1 + W2

Where W2 = 7.10×10⁻⁵J

So, W1 + W2 = W = K2

W1 + 7.10×10⁻⁵ = 4.25×10⁻⁵

W1 = 4.25×10⁻⁵ - 7.10×10⁻⁵

W = -2.85 * ×10⁻⁵ J

Work done by the electric force = -2.85 * ×10⁻⁵ J

b. What is the potential of the starting point with respect to the end point?

The change in potential energy is given as

W = ∆U

W = Q|V2 - V1| where V1 = 0 because the body starts from rest

So, W = QV2

Make V the Subject of the formula

V2 = W/Q

V2 = -2.85 * ×10⁻⁵ J / 9.40 nC

V2 = -2.85 * ×10⁻⁵ J / 9.40 * 10^-9C

V2 = −3031.9148936170212765957V

V2 = -3.03 * 10³ V

The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. What is the magnitude of the electric field?

The magnitude of the electric field is calculated as follows;

W = -Fd = -QEd

And E = V/d

E = -3.03 * 10³ V / 0.09 m

E = −33687.943262411347517730 V/m

E = -33.7kV/m

The magnitude of the electric field is 33.7kV/m

4 0

3 years ago