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geniusboy [140]
3 years ago
10

A car is running at a velocity of 50 miles/hour and the driver accelerates the car by 10miles/hour.How far the car travels from

this point in next 2 hours.If the acceleration is constant?l
Physics
2 answers:
dezoksy [38]3 years ago
4 0
Initial velocity u = 50 miles/hour
acceleration a = 10 miles/hour
Time t = 2 hours
Distance travelled S = ut + (at^2)/2
Substituting the values in the second equation of motion,
S = 50*2 + (10 * 2 *2)/2
S = 100 + 20
S = 120 miles
Therefore the distance travelled by the car in the next two hours is 120 miles
ELEN [110]3 years ago
3 0

Answer:

the distance travelled is 120 miles

Explanation:

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a desktop computer and monitor together draw about 2 A of current they plug into a wall outlet that is 120 V what is the Resista
Delvig [45]

Answer:

60 \Omega

Explanation:

the relation between current, voltage and resistance in an electrical circuit is given by Ohm's law:

V=IR

where V is the voltage, I is the current and R is the resistance. In this problem, the current is I=2 A, the voltage is V=120 V, therefore we can arrange the previous equation and find the resistance:

R=\frac{V}{I}=\frac{120 V}{2 A}=60 \Omega

7 0
3 years ago
K-2/5=11k literal equations
krek1111 [17]
-2/5 = 11k - k
-2/5 = 10k
-2/5/10 = k
-2/5 * 10 = k
-2/50 = k
k = -1/25.

-1/25 - 2/5 = 11k is true.
6 0
3 years ago
An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic
mezya [45]

Answer:

d. 332 V

Explanation:

Given;

number of turns in the wire, N = 40 turns

area of the coil, A = 0.06 m²

magnitude of the magnetic field, B = 0.4 T

frequency of the wave, f = 55 Hz

The maximum emf induced in the coil is given by;

E = NBAω

Where;

ω is angular velocity = 2πf

E = NBA(2πf)

E = 40 x 0.4 x 0.06 x (2 x π x 55)

E = 332 V

Therefore, the maximum induced emf in the coil is 332 V.

The correct option is "D"

d. 332 V

7 0
3 years ago
Given a force of 100 N and a acceleration of 5 m/s, what is the mass
SIZIF [17.4K]

Answer:

force = mass \times acceleration \\ 100 = m \times 5 \\ m =  \frac{100}{5}  \\ m = 20 \:  kg

8 0
2 years ago
Read 2 more answers
In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os
GalinKa [24]

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

I = Q.ω

ω =angular frequency

\omega=\dfrac{1}{\sqrt{CL}}

By putting the values

\omega=\dfrac{1}{\sqrt{CL}}

\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}

ω  = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33  A

I= 6.49 A

I = 6.5 A

E) 6.5 A

3 0
3 years ago
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