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2 years ago

10

A car is moving at 20 m/s. The driver steps on the gas pedal and accelerates the car for 4 seconds at an acceleration of 4 m/s2.

How fast is the car moving after it accelerates?

1 answer:

polet [3.4K]2 years ago

6 0

Answer:

The final velocity of the car is 36 m/s.

Explanation:

Given;

initial velocity of the car, u = 20 m/s

time of the car acceleration, t = 4 s

acceleration of the car, a = 4 m/s²

the final velocity of the car is calculated as;

v = u + at

where;

v is the final velocity of the car

v = 20 + (4 x 4)

v = 36 m/s

Therefore, the final velocity of the car is 36 m/s.

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A transformer consisting of two coils wrapped around an iron core is connected to a generator and a resistor (Resistor is connec

V125BC [204]

Answer:

The peak emf of the generator is 40.94 V.

Explanation:

Given that,

Number of turns in primary coil= 11

Number of turns in secondary coil= 18

Peak voltage = 67 V

We nee to calculate the peak emf

Using relation of number of turns and emf

$\dfrac{N_{1}}{N_{2}}=\dfrac{E_{1}}{E_{2}}$

$E_{1}=\dfrac{N_{1}}{N_{2}}\times E_{2}$

Where, N₁ = Number of turns in primary coil

N₂ = Number of turns in secondary coil

E₂ = emf across secondary coil

Put the value into the formula

$E_{1}=\dfrac{11}{18}\times67$

$E_{1}=40.94\ V$

Hence, The peak emf of the generator is 40.94 V.

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3 years ago

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2 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

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Divergent boundaries – where two plates are moving apart. ...

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3 years ago

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