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3 years ago

A car moves with the speed of 120m/s for 4 minutes ,calculate the distance covered by the car

Physics

1 answer:

Vsevolod [243]3 years ago

7 0

Answer:

960 m

Explanation:

Given that,

Speed = 120 m/s
Time taken = 4 minutes

We have to find the distance covered.

Firstly, let's convert time in seconds.

→ 1 minute = 60 seconds

→ 4 minutes = (4 × 60) seconds

→ 4 minutes = 240 seconds

Now, we know that,

→ Distance = Speed × Time

→ Distance = (4 × 240) m

→ Distance = 960 m

Therefore, distance covered is 960 m.

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3 years ago

Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau

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The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

Electric force,F = 9 × 10 ⁻⁹ N

Distance between charges,d = 7 × 10⁻⁴ m

Chrge,q₁ = q₂ =q C

From Columb's law;

$\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9} = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\ q = 7 \times 10^{-13} \ C$

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2 years ago

A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, $m_1 = 300\ g = 0.3\ kg$

Initial speed of the cart, $u_1=1.2\ m/s$

Mass of the larger cart, $m_2 = 2\ kg$

Initial speed of the larger cart, $u_2=0$

After the collision,

Final speed of the smaller cart, $v_1=-0.81\ m/s$ (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let $v_2$ is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_1v_1=m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}$

$v_2=0.301\ m/s$

So, the speed of the large cart after collision is 0.301 m/s.

4 0

3 years ago

a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe

Vika [28.1K]

Answer:

Explanation:

Given data

initital velocity u=15m/s

deceleration a=3m/s^2

time t= 4 seconds

final velocity v= ?

Applying the expression

v=u+at------1

substituting our data into the expression we have

v=15+3*4

v=15+12

v=27m/s

The velocity after 4 seconds is 27m/s

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3 years ago

A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$