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3 years ago

СРОЧНО ПОМОГИТЕ ПОЖАЛУЙСТА!!!!!!!!!!

Physics

1 answer:

Deffense [45]3 years ago

3 0

Боже, как это сложно! Ну ладно.

Между прочим это ты сам должен делать, а то не куда не поступишь!

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A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running

Katena32 [7]

Answer:

$t_a = 13.49 \ s$

The distance is $D = 55.2 \ m$

Explanation:

From the question we are told that

The maximum speed of the cheetah is $v = 103 \ km/h = 28.61 \ m/s$

The maximum of gazelle is $u = 76.5 \ km/h = 21.25 \ m/s$

The distance ahead is $d = 99.3 \ m$

Let $t_a$ denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

$v* t_a = d + u* t_a$

=> $28.61 t_a = 99.3 + 21.25t_a$

=> $7.36 t_a = 99.3$

=> $t_a = 13.49 \ s$

Now at t = 7.5 s

$7.5 v = D+ 7.5u$

=> $28.61 * 7.5 = D + 21.25* 7.5$

=> $7.36 * 7.5 =D$

=> $D = 55.2 \ m$

Hence the for the gazelle to escape the cheetah it must be 55.2 m

5 0

3 years ago

Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is

victus00 [196]

Answer:

3.15 N towards the positive x-axis

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = $\frac{-kQq}{r^2}$

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = $\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}$ = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = $\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}$ = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = 3.15 N towards the positive x-axis

3 0

3 years ago

______________ are chemicals in the body affecting how you feel.

7nadin3 [17]

Answer:

emotions

Explanation:

emotions are how you feel and can happen any time

Hope it helps <333

3 0

2 years ago

Read 2 more answers

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$