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2 years ago

7

a care starting from rest has an acceleration 0.3 m/s square, calculate the velocity and distance travelled by this car after 2

minutes?

1 answer:

Anna [14]2 years ago

3 0

Answer:

Final velocity (v) = 36 m/s

Distance traveled (s) = 2,160 m

Explanation:

Given:

Initial velocity (u) = 0

Acceleration (a) = 0.3 m/s

Time travel (t) = 2 minutes = 120 seconds

Find:

Final velocity (v) = ?

Distance traveled (s) = ?

Computation:

v = u + at

v = 0 + 0.3(120)

v = 0.3(120)

v = 36 m/s

Final velocity (v) = 36 m/s

Distance traveled (s) = ut + (1/2)at²

Distance traveled (s) = (0.5)(0.3 × 120 × 120)

Distance traveled (s) = 2,160 m

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The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which

spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒ $\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

On putting estimate values, we get

⇒ $\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}$

⇒ $\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}$

⇒ $v=1.94 \ cm$

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

$m=\frac{v}{u}$ and $m=\frac{h_{1}}{h_{0}}$

⇒ $\frac{v}{u} =\frac{h_{1}}{h_{0}}$

⇒ $\frac{1.94}{25}=\frac{{h_{1}}}{15}$

⇒ ${h_{1}}=1.16 \ mm$

8 0

2 years ago

A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta

serg [7]

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

$mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s$

So, the final velocity of the bullet is 9 m/s.

5 0

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.

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2 years ago

How much force does an 88kg astronaut exert on his chair while accelerating straight up at 10 m/s^2

Fed [463]

Force=mass*acceleration. So 88kg*10 m/s^2=880 newtons

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3 years ago

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pashok25 [27]

A right turn would be the answer probably.

8 0

3 years ago

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