Substances that protect body cells, including those of the immune system, from damage, are known as ____________

In the diagram, q1 and q2 are both +5.00*10^-8 C. What is the electric field at point P? Include a + or - sign to indicate the d

Damm [24]

Answer:

1389.1

Explanation:

6 0

2 years ago

According to the law of conservation of energy, how are potential energy and kinetic energy related?

frutty [35]

Your answer should be A, because every reaction has an equal opposite reaction

6 0

2 years ago

You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that

kogti [31]

Answer:

111.5 m

Explanation:

Given that You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.

Use first equation of motion

V = U - at

Since the car is going to rest, V = 0 and a = negative

0 = 14 - a × 0.5

0.5a = 14

a = 14 /0.5

a = 28 m/s^2

Let us use second equation of motion

S = Ut - 1/2at^2

S = 14 × 0.5 - 0.5 × 28 × 0.5^2

S = 7 - 3.5

S = 3.5 m

115 - 3.5 = 111.5

Therefore, you are 111.5 metres from the intersection (in m) when you begin to apply the brakes.

6 0

2 years ago

72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is locat

expeople1 [14]

Answer: $I=2 kg-m^2$

Explanation:

Given

mass $m=72 kg$

Force $F=5 N$

door knob is located at a distance of r=0.8 m from axis

Angular acceleration of door $\alpha =2 rad/s^2$

Torque $T=I\alpha =F\times r$

where I=moment of inertia

$5\times 0.8=I\times 2$

$I=2 kg-m^2$

4 0

3 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

2 years ago

Substances that protect body cells, including those of the immune system, from damage, are known as _____________.