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marta [7]
3 years ago
10

What is the BEST example of a purebred

Physics
1 answer:
ki77a [65]3 years ago
8 0

no se   nada pvodebrhtre

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A daredevil is shot out of a cannon at 45.0° to the horizontal with an initial speed of 31.0 m/s. A net is positioned a horizont
hodyreva [135]

Answer:

s = vcos(x)t

50 = 25cos(45)t

cos(45)t = 2

t = 2/cos(45) = 2sqrt(2)

h = vsin(x)t + gt^2/2

h = 25sin(45)*2sqrt(2) - 4.9*8

h = 10.8 metres

Explanation:

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Which type of force can be in the same it opposite direction?
LuckyWell [14K]

Answer:

C) unbalanced

Explanation:

Equal forces acting in opposite directions are called balanced forces. Balanced forces acting on an object will not change the object's motion. When you add equal forces in opposite direction, the net force is zero.

4 0
3 years ago
You have a series circuit powered by a 9V battery. If you double the amount of
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It gets 2 times weaker
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How can someone, like myself, share roughly 50% of their DNA with a 1st Cousin?
iragen [17]

Answer:

that's your 1st cousin which is your mom or dad's sibling's child sooo when you're that close and that related i would say that that would be the reason why.

Explanation:

hoped that helped!!

4 0
3 years ago
A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4
HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

s = ut+\frac{1}{2}at^2

2.8 \text{ m} = 0+\dfrac{gt^2}{2}

t = \sqrt{\dfrac{5.6}{g}}

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g                   </em> (It is going up)

Using the equation of motion

s = ut+\frac{1}{2}at^2

1.2\text{ m} = ut-\frac{1}{2}gt^2

Substituting the value of <em>t</em> from the previous equation,

1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}

u\sqrt{\dfrac{5.6}{g}} = 4.0

Taking <em>g</em> = 9.8 m/s²,

u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}

The initial velocity is

v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}

4 0
4 years ago
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