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Irina18 [472]
3 years ago
10

A car has a speed of 20 m/s. If the speed of the car increases to 30 m/s in 5 seconds, what is the car’s acceleration?

Physics
1 answer:
Tema [17]3 years ago
7 0

2

Explanation:

2 times 5 is 10/ 20+10=30

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two girls drag a bag across the floor. One girl exerts a force of 10 newtons and the other girl a force of 30 Newton's in the sa
lara [203]

Answer:

Explanation:

The resultant would be a combination of the two forces. We need to add them together since they are exerting a force in the same direction.

10+30= 40 N of force

5 0
3 years ago
If the number of homes with a pet dog is equal to 250, how many total homes are repersented by the chart?
scoray [572]
500 because 250 x 2 = 500
3 0
3 years ago
In the visible spectra of stars, absorption lines of hydrogen are produced when atoms are excited from n = 2 to higher levels (t
grigory [225]

Answer:

3. relatively high temperature, about 10,000 K, so that significant numbers of electrons are excited from the ground state, n = 1, to the first excited state, n = 2, but not too many of them have been ejected completely from the atoms

Explanation:

If hydrogen absorption lines are very strong in the visible spectrum of a particular star that means the population of electron in n = 2 is very high so on being exited they absorb radiation in Balmer series and give rise to absorption spectrum. The average temperature required to excite electron in hydrogen atom from n=1 to n = 2   is 10000K .

4 0
3 years ago
A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 x 10–3 kg∙m2 is suspended from the ceiling. A rope passes over
vekshin1

Answer:

22J

Explanation:

Given :

radius 'r'= 3cm

rotational inertia 'I'=4.5 x 10^{-3} kgm²

mass on one side of rope 'm_{1'= 2kg

mass on other side of rope'm_{2' =4kg

velocity'v' of mass m_{2' = 2m/s

Angular velocity of the pulley is given by

‎ω = v /r => 2/ 3x 10^{-2

‎ω = 66.67 rad/s

For the rotating body, we have

KE = \frac{1}{2} I ω²

KE_p = \frac{1}{2} (4.5 *10^{-3} )(66.67^{2} )

KE_p = 10J

Next is to calculate kinetic energy of the blocks :

KE_{b} = \frac{1}{2} (m_1 + m_2).v^2\\KE_b= \frac{1}{2} (2+4).2^2

KE_b=12J

Therefore, the total kinetic energy will be

KE = KE_p + KE_b =10 + 12

KE= 22J

6 0
3 years ago
A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle
NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

Power,P=(F_d+f)\times v'

340=(4.1+4.635) v'=8.735v'

v'=\frac{340}{8.735}=38.9m/s

v'=87.1 mph

1 m=0.00062137 miles

1 hour=3600 s

7 0
3 years ago
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