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3 years ago

What is the energy equivalent of an object with a mass of 4.1 kg?

Physics

2 answers:

vichka [17]3 years ago

7 0

Answer:

3.7 × 1017 J

Explanation:

siniylev [52]3 years ago

6 0

To determine the energy equivalent of an object, we use the famous equation of Einstein which is E=mc^2 where m is the mass of the object and c is the speed of light (3x10^8 m/s). We calculate as follows:

E = mc^2
E = 4.1 kg (3x10^8 m/s)^2
E = 3.69x10^17 J

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A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at

docker41 [41]

Answer:

$y = 4.36$

Explanation:

Let the height of the ladder be L

$L = 10$

Also:

Let $x = distance\ from\ the\ base\ of\ the\ ladder$

Let $y = height\ of\ the\ base\ of\ the\ ladder$

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

$L^2 = x^2 + y^2$ --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

$x = 9$

Substitute 9 for x and 10 for L in $L^2 = x^2 + y^2$

$10^2 = 9^2 + y^2$

$100 = 81 + y^2$

Make $y^2$ the subject

$y^2 = 100 - 81$

$y^2 = 19$

Make y the subject

$y = \sqrt{19$

$y = 4.36$

<em>Hence, the true distance at that point is approximately 4.36ft</em>

8 0

2 years ago

A 0.49-kg cord is stretched between two supports,7.3m apart. When one support is struck by a hammer, a transverse wave travels d

Wewaii [24]

Answer:

T= 5.18N

Explanation:

u = mass of chord / length of chord

u = 0.49/ 7.3

u = 0.067 kg/m

Velocity of sound waves (v) =length of chord / time taken for wave to travel

v = 7.3 / 0.83 = 8.795m/s

Tension is calculated below using the formula

T = v² * u

T = (8.795)² x 0.067

T= 5.18N

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2 years ago

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Pepsi [2]

Answer:

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Explanation:

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nikdorinn [45]

Answer:

Also 3s.

Explanation:

Each component is independent in two dimensional motion. This means that <em>how much time does something take to reach the ground when dropped is independent from any horizontal velocity</em>. If at one run a drop lasts 3s, at another run with twice the (horizontal) velocity and same height will also last 3s, no matter what.

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2 years ago

A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj

Norma-Jean [14]

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

Time to fall from zero speed to ground = rise time = 1.76 sec

Vx = V cos 35 = 24.6 m / sec horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

8 0

2 years ago