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4 years ago

Isaiah has samples of two different elements, one metal and one nonmetal.

Physics

1 answer:

Brums [2.3K]4 years ago

5 0

Answer:

sample B is non metal

............

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g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron

Butoxors [25]

Answer:

The value is $v = 2.999 *10^{8} \ m/s$

Explanation:

From the question we are told that

The time taken to travel to the planet from earth is $t = 189 \ light-years$

The time to be spent on the ship is $t_{s} = 12 \ years$

Generally speed can be obtained using the mathematical relation represented below

$t_s = 2 * t * \sqrt{1 - \frac{v^2}{c^2 } }$

The 2 in the equation show that the trip is a round trip i.e going and coming back

=> $12 = 2 * 189 * \sqrt{1 - \frac{v^2}{(3.0*10^{8})^2 } }$

=> $v = 2.999 *10^{8} \ m/s$

5 0

3 years ago

True or false? Thanks,

vichka [17]

Answer: The answer is true

7 0

1 year ago

Read 2 more answers

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago

A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t

navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that $v^2=u^2+2as$

$16^2=24^2+2\times a\times 50$

$a=-3.2m/sec^2$

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So $v^2=u^2+2as$

$0^2=24^2-2\times 3.2\times s$

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration $a=-3.2m/sec^2$

So time $t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec$

7 0

3 years ago

Please answer fast in hindi

Ipatiy [6.2K]

Answer:

1. Dheere Dheere (slowly slowly)

2. Har (every)

3. Kal (tomorrow)

4. Mat (don't)

5. Andar (inside)

sorry I wasn't able to write in hindi

8 0

3 years ago