All categories

3 years ago

A bus travelling 30.0 km/h east has a constant increase in speed of 1.5 m/s. What is its velocity 6.8s later

Physics

1 answer:

alekssr [168]3 years ago

7 0

Answer:

8.72 seconds before

Explanation:

and 27.1 secs after

You might be interested in

Why is the contribution of the wavelets lying on the back of secondary wave front zero?

Tresset [83]

Answer:

The contribution of the wavelets lying on the back of the wave front is zero because of something known as the Obliquity Factor. It is assumed that the amplitude of the secondary wavelets is not independent of the direction of propagation, Sources: byju's.com

3 0

2 years ago

Read 2 more answers

1. Place these unknown pH test papers in order from most acidic to most alkaline.

Naddika [18.5K]

Answer: B

Explanation: look at the chart, easy

7 0

2 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

2 years ago

Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and

dimaraw [331]

Answer:-q

Explanation:

Given

Capacitor is charged to a battery and capacitor acquired a charge of q i.e.

+q on Positive Plate and -q on negative Plate.

If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by

$c=\epsilon_0 \cdot \frac{A}{d}$

where A=area of capacitor plate

d=Separation between plates

This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q

8 0

3 years ago

Which country had the largest population in 1997

nignag [31]

China i hope this helped

6 0

3 years ago