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3 years ago

A 845kg dragster (very fast car) accelerates from 2m/s to 30m/s in 0.9s. Determine the average force exerted on the dragster.

Physics

1 answer:

Stella [2.4K]3 years ago

7 0

Answer:

Explanation:

Just use the Force formula.

F = M . A

Acceleration Formula

A = V - Vo / Time

So...

F = 845 . (30 - 2 / 0.9)

F = 845 . 20

F = 16900 N

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Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0

3 years ago

Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r

IrinaK [193]

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V is the potential difference

Calling $C_1$ the capacitance of capacitor 1 and $V_1$ its potential difference, the energy stored in capacitor 1 is

$U=\frac{1}{2}C_1 V_1^2$

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: $C_2 = \frac{C_1}{2}$

- The voltage is twice the voltage of capacitor 1: $V_2 = 2 V_1$

so the energy stored in capacitor 2 is

$U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2$

So the ratio between the two energies is

$\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}$

4 0

3 years ago

Can someone please help me! BRAINLIEST ANSWER! What happens after the Gulf Stream travels north?

Irina-Kira [14]

A should be the answer since it makes the water down there cold and the air also. (asked my grandma haha

3 0

3 years ago

A helium–neon laser emits a beam of circular cross section with a radius r and a power P.?. (a) Find the maximum electric field

olchik [2.2K]

there is a relation between intensity of light beam and the magnitude of electric field.I=(1/2)cϵonE2=P/πr2 E2=2P/cϵonπr2 E= magnitude of electric field n= refractive index of medium μoϵ0=1/c2
energy= power*time = P*(1m/speed of light)energy=(P∗1m)/c

5 0

4 years ago

A water molecule perpendicular to an electric field has 1.40×10−21 J more potential energy than a water molecule aligned with th

vesna_86 [32]

Answer:

The strength of the electric field is 2.258 x 10⁸ N/C

Explanation:

Potential energy = -p*ECosθ

where;

d is the dipole moment

E is the electric field

θ is the angle of inclination

→When the water molecules is perpendicular to the field, θ =90° and potential energy = p*ECosθ = d*ECos0 = 0

Total potential energy = 0 + 1.40×10⁻²¹J, since it is 1.40×10⁻²¹J more

→When the water molecules is aligned to the field, θ =0°

potential energy = -p*ECosθ

dipole moment, p = 6.2×10⁻³⁰Cm

= -(6.2×10⁻³⁰)*(E) Cos0

0 + 1.40×10⁻²¹ J = -(6.2×10⁻³⁰)*(E),

E = (1.40×10⁻²¹ J)/(6.2×10⁻³⁰)

E = 2.258 x 10⁸ N/C

Therefore, the strength of the electric field is 2.258 x 10⁸ N/C

3 0

3 years ago