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3 years ago

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a man is 9 m behind a dorr of train when it starts moving with a=2ms^-1. how far the man have to run and after what time will he

catch the train. what is his full speed

1 answer:

Naya [18.7K]3 years ago

8 0

Is their a multiple choice to choose from I'm not sure the answer I got is even right.
That would be very helpful.

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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo

yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is = $102^{\circ}-65.3^{\circ}=36.7^{\circ}$

Speed of first plane = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, $r^2$ representing the distance between the planes, we see that:

$r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295$

r = 741.6959 m

3 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

What is the conclusion of coin and feather experiment?

valina [46]

Answer:

So the conclusion is that in presence of air net force acting downward reduces for feather and hence falls slower than coin. But in absence of air resistance, net downward force is just equal to force due to gravity which is same for both coin and feather and hence they fall down at the same rate.

5 0

3 years ago

Can someone help me?

Setler [38]

Answer: (1, 30), (2,10), (3,40), (4,20)

Explanation:

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2 years ago

Gardeners understand that plants have specific pH requirements. Most plants thrive in soil with a pH that is slightly acidic, ne

Delvig [45]

Answer:

D) Add a base like limestone or calcium carbonate.

Explanation:

-To reduce the soil's pH, you need to add an alkaline or base solution.

-Calcium carbonate has an alkanizing property and forms a carbonic acid and calcium hydroxide when dissolved in water.

-The solution will reduces the soil's pH.

8 0

3 years ago