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3 years ago

Altitude means the what?

Physics

1 answer:

andrew-mc [135]3 years ago

7 0

Answer:

the height of a object or point relation to sea level or ground level

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Tom Cruise jumped from one building to other building while filming the roof chase scene in Mission: Impossible - Fallout. He di

OLga [1]

Answer:

$v_0=9.9\ m.s^{-1}$

Explanation:

Given:

angle of launch of projectile from horizontal, $\theta=15^{\circ}$

range of projectile, $R=5\ m$

We have formula for the range of projectile:

$R=\frac{v_0^2\times sin\ 2\theta}{g}$

putting the respective values

$5=\frac{v_0^2\times sin\ 30^{\circ}}{9.8}$

$v_0=9.9\ m.s^{-1}$ is the velocity with which Tom should jump to land on the other roof.

3 0

3 years ago

In an arcade game a 0.099 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releas

Sonja [21]

Answer:

The speed of disk is 1.98 $\frac{m}{s}$

Explanation:

Given:

Mass of $m = 0.099$ kg

Spring constant $k = 244 \frac{N}{m}$

Compression of spring $x = 4 \times 10^{-2}$ m

From energy conservation theorem,

Spring potential energy converted into kinetic energy,

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

$v = \sqrt{\frac{k x^{2} }{m} }$

$v = \sqrt{\frac{244 \times 16 \times 10^{-4} }{0.099} }$

$v = 1.98 \frac{m}{s}$

Therefore, the speed of disk is 1.98 $\frac{m}{s}$

8 0

3 years ago

What is a analogy for transition metals

Ganezh [65]

A simple metallic band model is proposed for the transition metal mono antimonides, by analogy to the transition metals.

6 0

3 years ago

Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of

slamgirl [31]

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

Explanation:

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

$v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2$

Substituting the velocities in the equation

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0

3 years ago

A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of

forsale [732]

Impulse = Force * times and also Impulse = change in momentum.

Given that the mass does not change, change if momentum = mass * (final velocity - initial velocity)

Given that you know mass and initial velocity (which is the velicity before the cart hits the wall) you need the final velocity (which is the velocity after the cart hits the wall).

Answer: the velocity of the cart after it hits the wall.

6 0

3 years ago