All categories

2 years ago

.

Physics

2 answers:

Lerok [7]2 years ago

5 0

Answer:

Explanation:

48km:1hr (10sec×60min=600hr)

¥ :600hr

1hr÷¥=1. 48×600÷1=2800

2800÷60=47

47 will be the acceleration

Leni [432]2 years ago

3 0

$\text{Initial velocity,}~ u = 0 ~ ms^{-1}\\\\\text{Final velocity,}~ v = 48~kmh^{-1} = \dfrac{48\times 1000}{3600} ms^{-1} = 13.33 ms^{-1}\\ \\\text{Time,}~ t = 10~ sec\\\\\text{Acceleration,}~ a = \dfrac{v-u}{t} = \dfrac{13.33-0}{10} = \dfrac{13.33}{10} = 1.33 ~ ms^{-2}\\\\\text{Hence the acceleration of the car is}~ 1.33 ~ms^{-2}.$

You might be interested in

WHICH IS THE HARDEST SUBSTANCE IN EARTH?

PIT_PIT [208]

Diamond is the hardest naturally occurring substance found on Earth.

8 0

3 years ago

Read 2 more answers

Why is it more helpful to know a tornadoes velocity rather than its speed

victus00 [196]

So you not only know<span> how fast it is going, but where it is going.</span>

5 0

3 years ago

Read 2 more answers

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr

aleksandr82 [10.1K]

Answer:

Explanation:

We shall apply Doppler's effect to solve the problem .

Formula for apparent frequency for a source of sound approaching an observer is as follows .

f₁ = f₀ V / (V - v )

where f₁ and f₀ are apparent and real frequency of source , V and v is velocity of sound and velocity of approaching source respectively .

Putting the given values and knowing that speed of sound is 340 m /s

f₁ =346x 340 / (340 - 39.6 )

f₁ = 391.6 Hz

In case of receding train , the formula is

f₂ = f₀ V / (V + v )

Putting the values

f₂ = 346x 340 / (340 + 39.6 )

= 309.9 Hz

Change in frequency = 391.6 - 309.9

= 81.7 Hz .

4 0

3 years ago

How does a rollercoaster convert potential energy into kinetic energy and then back to potential energy again and again?

Vladimir79 [104]

When a roller coster goes down hill all its stored energy is bieng used as electricity is bieng converted into energy (moving)

5 0

3 years ago

A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The

malfutka [58]

Answer:

$\theta = 23.32^o$

$\mu_s = 0.27$

$s = 0.948 \ m$

Explanation:

From the question we are told that

The mass of the bag is $m_b = 25.0 \ kg$

The normal force experienced is $F_n = 225 \ N$

The maximum acceleration of the bag is $a = 2.40 \ m/s^2$

Generally this normal force experience by the bag is mathematically represented as

$F_n = mg cos \theta$

=> $225 = (25 * 9.8) cos \theta$

=> $0.9183 = cos \theta$

=> $\theta = cos^{-1}[0.9183]$

=> $\theta = 23.32^o$

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

$F_f = F_s$

Hence $F_f$ is mathematically represented as

$F_f = \mu_s * F_n$

While $F_s$ is mathematically represented as

$F_s = m * a$

$\mu_s * F_n = m * a$

=> $\mu_s * 225 = 25 * 2.40$

=> $\mu_s = 0.27$

Generally from the workdone equation we have that

$KE_f - KE_i = W_f$

Here $W_f$ is the work done by friction which is mathematically represented as

$W_f = m * g * \mu_k * s$

Here s is the distance covered by the bag

$KE_f$ is zero given that velocity at rest is zero

and

$KE_i = \frac{1}{2} * m* v_i^2$

$\frac{1}{2} * m* v_i^2 = m * g * \mu_k * s$

=> $\frac{1}{2} * v_i^2 = g * \mu_k * s$

substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that

$\frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s$

=> $s = 0.948 \ m$

4 0

4 years ago