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3 years ago

a car headed north at 15 m/s accelerates for 4.25 s to reach a velocity of 28.3 m/s. What is the acceleration of the car?

Physics

1 answer:

Liono4ka [1.6K]3 years ago

7 0

<u>Answer:</u>

The acceleration of the car is $3.13 m/s^2$

<u>Explanation:</u>

In the question it is given that car initially heads north with a velocity $15 m/s$ . It then accelerates for $4.25 s$ and in the end its velocity is $28.3 m/s$ .

initial velocity $u = 15 m/s$

time $t=4.25 s$

final velocity $v=28.3 m/s$

The equation of acceleration is

$a= \frac{(v-u)}{t}$

$= \frac{(28.3-15)}{4.25} = \frac {13.3}{4.25} =3.13m/s^2$

The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.

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Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in se

just olya [345]

Answer:

1.61ohms and 4.39ohms

Explanation:

According to ohm's law which States that the current (I) passing through a metallic conductor at constant temperature is directly proportional to the potential difference (V) across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the current

Rt is the effective resistance

Let the resistances be R and r

When the resistors are connected in series to a 12.0-V battery and the current from the battery is 2.00 A, the equation becomes;

12 = 2(R+r)

Rt = R+r (connection in series)

6 = R+r ...(1)

If the resistors are connected in parallel to the battery and the total current from the battery is 10.2 A, the equation will become;

12 = 10.2(1/R+1/r)

Since 1/Rt = 1/R+1/r (parallel connection)

Rt = R×r/R+r

12 = 10.2(Rr/R+r)

12(R+r) = 10.2Rr ... (2)

Solving equation 1 and 2 simultaneously to get the resistances. From (1), R = 6-r...(3)

Substituting equation 3 into 2 we have;

12{(6-r)+r} = 10.2(6-r)r

12(6-r+r) = 10.2(6r-r²)

72 = 10.2(6r-r²)

36 = 5.1(6r-r²)

36 = 30.6r-5.1r²

5.1r²-30.6r +36 =

r = 30.6±√30.6²-4(5.1)(36)/2(5.1)

r = 30.6±√936.36-734.4/10.2

r = 30.6±√201.96/10.2

r = 30.6±14.2/10.2

r = 44.8/10.2 and r = 16.4/10.2

r = 4.39 and 1.61ohms

Since R+r = 6

R+1.61 = 6

R = 6-1.61

R = 4.39ohms

Therefore the resistances are 1.61ohms and 4.39ohms

5 0

3 years ago

A yoyo with a mass of m = 150 g is released from rest as shown in the figure.

avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

I is moment of inertia
α is angular acceleration
T is tension in the rope
r is inner radius
R is outer radius
f is frictional force

rT - Rf = Iα ----- (1)

T - f = Ma -------- (2)

a = Rα

where;

a is the linear acceleration of the yoyo

Torque equation for frictional force;

$f = (\frac{r}{R} T) - (\frac{I}{R^2} )a$

solve (1) and (2)

$a = \frac{TR(R - r)}{I + MR^2}$

since the yoyo is pulled in vertical direction, T = mg $a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2$