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Dmitrij [34]
3 years ago
14

Please answer. True or false and how do you know?

Physics
2 answers:
maks197457 [2]3 years ago
7 0
Answer: False

We know this as the acceleration of the ball would be dependent on the force used to throw it up.

Hope this helped!
blagie [28]3 years ago
4 0

Answer:False

Explanation:

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A tennis ball is dropped from 1.62 m above the
natita [175]

Answer:

-5.63 m/s

Explanation:

Given:

y₀ = 1.62 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.62 m)

v = -5.63 m/s

8 0
3 years ago
A 195 g glider is moving at 5.3 m/s on a frictionless air track. It then collides with a stationary 295 g glider.
lord [1]

Answer:For a perfectly elastic collision, the final velocities of the carts will each be 1/2 the velocity of the initial velocity of the moving cart. For a perfectly inelastic collision, the final velocity of the cart system will be 1/2 the initial velocity of the moving cart.

Explanation:

3 0
3 years ago
A 100-m-long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.050 m
solong [7]

Answer:

1.6 x 10^-5 T

Explanation:

i = 4 A

r = 0.05 m

The magnetic field due to long wire at a distance r is given by

B = \frac{\mu _{0}}{4\pi }\times \frac{2i}{r}

B = 10^-7 x 2 x 4 / 0.05

B = 1.6 x 10^-5 T

3 0
4 years ago
Does anyone know this or done it? It’s for life skills and about MyPlate:)
vivado [14]
The answer is no I’ve not done that
6 0
3 years ago
a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

4 0
3 years ago
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