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2 years ago

A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

Physics

1 answer:

BaLLatris [955]2 years ago

8 0

The answer would be 2.8m height on earth takes
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>

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a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

The headlights of a car emit light of wavelength 400 nm and are separated by 1.2 m. The headlights are viewed by an observer who

densk [106]

Answer:

The most correct option is;

B. 10 km

Explanation:

$L = \frac{y \times d}{1.22 \times \lambda} = \frac{1.2 \times 0.004}{1.22 \times 400 \times 10^{-9}} = 9836.066 \ km$

Where:

y = Distance between the two headlights

d = Aperture of observers eye

λ = Wavelength of light

L = Distance between the observer and the headlight

Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.

Also we have

sinθ = y/L = 1.22 (λ/d)

$= 1.22 \times \frac{400 \times 10^{-9}}{0.004}$

sinθ = 1.22×10⁻⁴ rad

6 0

2 years ago

Read 2 more answers

The Palo Verde nuclear power generator of Arizona has three reactors that have a combined generating capacity of 3.937×109 W . H

den301095 [7]

Answer:

t = 2.68 x 10¹⁴ years

Explanation:

First we need to find the amount of energy that Sun produce in one day.

Energy = Power * Time

Energy of Sun in 1 day = (3.839 x 10²⁶ W)(1 day)(24 hr/1 day)(3600 s/ 1 hr)

Energy of Sun in 1 day = 3.32 x 10³¹ J

Now, the time required by the nuclear power generator, in years, will be:

Energy of power generator = Energy Sun in 1 day = 3.32 x 10³¹ J

3.32 x 10³¹ J = Power * Time

3.32 x 10³¹ J = (3.937 x 10⁹ W)(t years)(365 days/1 year)(24 hr/1 day)(3600 s/ 1 hr)

t = 3.32 x 10³¹ /1.24 x 10¹⁷

<u>t = 2.68 x 10¹⁴ years</u>

8 0

2 years ago

a motorist traveling at 18m/s approaches traffic lights when he is 30 m from the stop line they turn red it takes 0.7 s before h

Gelneren [198K]

Answer:

35.2m

Explanation:

S= V^2 - U^2 ÷ 2A

S= (0) - (18) ÷ 2(4.6)

S= 324 ÷ 92

S= 35.2m

7 0

2 years ago

In which applications are sound waves reflected? Check all that apply.

mojhsa [17]

detecting diseases, locating damaged cells, monitoring fetal development

8 0

2 years ago

Read 2 more answers