3 years ago

0.2 points

Physics

1 answer:

Nutka1998 [239]3 years ago

7 0

The answer is 2.6m/s2

You might be interested in

Asap

Elan Coil [88]

Answer:

sorry i don't understand

Explanation:

3 0

3 years ago

Read 2 more answers

A biker to move from speed of 0 m/s to final speed of 25 m/s in 10 s. What is the acceleration of the bike?

Rzqust [24]

Answer:

a = 2.5 m/s²

Explanation:

given,

initial speed = 0 m/s

final speed = 25 m/s

time = 10 s

acceleration of bike is equal to change in velocity per unit time

$a = \dfrac{\Delta v}{t}$

$a = \dfrac{v_f-v_i}{t}$

$a = \dfrac{25-0}{10}$

a = 2.5 m/s²

hence, acceleration of bike is equal to a = 2.5 m/s²

4 0

3 years ago

A chemical property is a change in _____. A density B composition of matter C hardness D physical state

NemiM [27]

Composition of matter

5 0

3 years ago

Read 2 more answers

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

djverab [1.8K]

Answer:

0.005 V

Explanation:

We are given that

Initial circumference of circular loop=C=165 cm

Rate of circumference, $\frac{dC}{dt}=12 cm/s$

Magnetic field,B=0.5 T

We have to find the induced emf at time t=9 s

We know that induced amf,E= $\frac{Bd(A)}{dt}$

Area of circular coil,A= $\pi r^2$

$E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}$

Circumference of circular coil,C= $2\pi r$

$165=2\pi r$

$r=\frac{165}{2\pi}$

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}\times (12)=\frac{6}{\pi} cm/s=\frac{6\times 10^{-2}}{\pi} m/s$

Radius of coil at time t=9 s

$r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m$

$1 m=100 cm$

E= $-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V$

Magnitude of induced emf=0.005 V

4 0

3 years ago

Read 2 more answers

How much heat is absorbed from a 56.00 g sample of Mercury when its tempreature change is 289K?

AveGali [126]

Heat required to raise the temperature of mercury is given as

$Q = ms\Delta T$

here given that

m = 56 g

s = specific heat capacity of mercury = 0.140 J/g C

$\Deta T = 289 k$

now here we have

$Q = 56 * 0.140 * 289$

$Q = 2265.76 J$

so it required 2265.76 J of heat

8 0

3 years ago