Ask question

Ask question

All categories

3 years ago

8

Arace car starts from rest and accelerates uniformly to a speed of 40 m/s in 8.

1 answer:

9966 [12]3 years ago

8 0

Answer:

160 m

Explanation:

For computing the distance first we have to determine the acceleration which is shown below:

v = u + at

where,

v = 0 = initial velocity

u = 40 = uniformly to a speed

t = 8 = time

So, the a is

= 40 ÷ 8

= 5 m/s

Now we have to use the distance formula i.e

d = ut + 1 ÷ 2 at^2

= 1 ÷2 × 5 × 8^2

= 160 m

Basically we have to applied the above formulas so that the distance could come

You might be interested in

If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t

solong [7]

Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s

3 0

3 years ago

The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t

hodyreva [135]

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

$f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}$ .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

$f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}$ .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

7 0

3 years ago

An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 5.00 V. T

Maksim231197 [3]

Answer:

35

Explanation:

We are given that

Initial voltage, $V_1=120 V$

Final voltage, $V_2=5 V$

Number of tuns in primary coil of the transformer, $N_p=840$

Rms current, $I_{rms}=580mA=580\times 10^{-3} A$

$1 mA=10^{-3} A$

We have to find the number of turns are there on the secondary coil.

We know that

$\frac{N_s}{N_p}=\frac{V_2}{V_1}$

Using the formula

$\frac{N_s}{840}=\frac{5}{120}$

$N_s=\frac{5}{120}\times 840=35$

Hence, there are number of turns on the secondary coil=35

8 0

3 years ago

A body travels at an initial speed of 2.5 m/s. Given a constant acceleration of 0.2 m/s 2 what is the speed of the body at time

garri49 [273]

Answer:

<u>We are given:</u>

u = 2.5 m/s

a = 0.2 m/s/s

t = 25 seconds

v = v m/s

<u>Solving for 'v':</u>

From the first equation of motion:

v = u + at

Replacing the values

v = 2.5 + (0.2)(25)

v = 2.5 + 5

v = 7.5 m/s

6 0

3 years ago

A 2,000 kg railroad car is coasting on a track at a constant velocity of 20 m/s. As the car coasts under a loading ramp, a 500 k

BlackZzzverrR [31]

Answer:

the car with the hay should slow to 16m/s if the bale of hay is dropped into it.

3 0

3 years ago