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Marysya12 [62]
2 years ago
13

A ball rolls off an 8.0 m high building and strikes the ground 5.0 m away from the base of the building. How fast was the ball r

olling?
Physics
1 answer:
HACTEHA [7]2 years ago
4 0

Answer: horizontal speed is 3.9 m/s

Explanation: when ball starts to drop, its vertical speed v0 is zero.

We can calculate dropping time from s = v0t +0.5gt².

Dropping time t= √(2s/g)= √((2·8.0 m)/9.81 m/s²)= 1.277 s

Because ball travels horizontal distance s= 5.0 m

HorizontalSpeed v = s/t = 5.0 m/1.277s= 3,915 m/s

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Serggg [28]
You don't need to do any calculations, you are already given that n=1.5
n is index of refraction. it's 1.5
4 0
2 years ago
Which two staments explain how a cell's parts help it get nutrients
Goryan [66]

Answer:

I think it's A and D.

6 0
2 years ago
Does electrons have more mass than neutrons?
musickatia [10]

Answer:

No, neutrons have about the same mass as a proton, but both have more mass than electrons.

Hope this helps a bit,

Flips

3 0
2 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T
monitta

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

     d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

     d = 1/600

     d = 1.67 10⁻³ mm

    d = 1-67 10⁻³ m

Let's calculate the angle

    sin θ = m λ / d

    θ  = sin⁻¹ (m λ / d)

First order

    θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

    θ₁ = sin⁻¹ (3.93 10⁻¹)

    θ₁ = 23.14 °

Second order

     θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

     θ₂ = sin⁻¹ (0.786)

     θ₂ = 51.81 °

3 0
3 years ago
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