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3 years ago

15

If someone drops a cup, it falls to the ground. Why doesn't the gravitational force between the person's hand and the cup keep t

he cup from falling?

1 answer:

Genrish500 [490]3 years ago

5 0

Answer:

The gravity from the person's hand is weaker than the gravity from the pull of the earth

Explanation:

The gravity from the person's hand is weaker than the gravity from the pull of the earth

You might be interested in

What is the wavelength of a monochromatic light beam, where the photon energy is 2.70 × 10^−19 J? (h = 6.63 ×10^−34 J⋅s, c = 3.0

SOVA2 [1]

Answer:

Wavelength = 736.67 nm

Explanation:

Given

Energy of the photon = 2.70 × 10⁻¹⁹ J

Considering:

$Energy=h\times frequency$

where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

So, Formula for energy:

$Energy=h\times \frac {c}{\lambda}$

Energy = 2.70 × 10⁻¹⁹ J

c = 3×10⁸ m/s

h = 6.63 x 10⁻³⁴ J.s

Thus, applying in the formula:

$2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}$

Wavelength = 736.67 × 10⁻⁹ m

1 nm = 10⁻⁹ m

So,

<u>Wavelength = 736.67 nm</u>

8 0

3 years ago

Finish the sentence below with the word that makes the sentence true.

Keith_Richards [23]

Answer:

Animals must eat other plants or animals to obtain the <u>energy</u> stored in the food

Explanation:

One classification of living organisms, according to the way they obtain energy, is that of autotrophs and heterotrophs. The first group is represented by plants, which process their own nutrients from inorganic matter.

<u>Animals -heterotrophes- are unable to process their own nutrients</u>, so they must obtain them from other organisms, either plants or animals. External food sources provide them with nutrients, which contain the energy substrate needed to perform their vital functions.

Learn more:

Autotrophs and heterotrophs brainly.com/question/7695115

7 0

3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

3 years ago

A 12.0-g bullet is fi red horizontally into a 100-g wooden block initially at rest on a horizontal surface. After impact, the bl

allochka39001 [22]

Answer:

$v_1 = 91.3 m/s$

Explanation:

By energy conservation and work energy theorem we can say that after bullet hits the block, it will move on the rough floor and comes to rest

so here work done by frictional force = change in kinetic energy

so we know that

$W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$-\mu mg d = \frac{1}{2}m(v_f^2 - v_i^2)$

$-0.650(9.81)(7.5) = \frac{1}{2}(0 - v_i^2)$

$47.8 = \frac{1}{2}v^2$

$v = 9.78 m/s$

now by momentum conservation we have

$m_1 v_1 = (m_1 + m_2) v$

$12 v_1 = (100 + 12) 9.78$

$v_1 = 91.3 m/s$

3 0

2 years ago

In a certain two slit diffraction experiment, two slits 0.02mm wide are spaced0.2mm between centers.(a) How many fringes appear

Kamila [148]

Answer:

a) m = 10 and b) λ = 3.119 10⁻⁷ m

Explanation:

In the diffraction experiments the maximums appear due to the interference phenomenon modulated by the envelope of the diffraction phenomenon, for which to find the number of lines within the maximum diffraction center we must relate the equations of the two phenomena.

Interference equation d sin θ = m λ

Diffraction equation a sin θ = n λ

Where d is the width between slits (d = 0.2 mm), a is the width of each slit (a = 0.02 mm). θ is the angle, λ the wavelength, m and n are an integer.

Let's find the relationship of these two equations

d sin θ / a sin θ = m Lam / n Lam

The first maximum diffraction (envelope) occurs for n = 1, let's simplify

d / a = m

Let's calculate

m = 0.2 / 0.02

m = 10

This means that 10 interference lines appear within the first maximum diffraction.

b) let's use the interference equation, remember that the angles must be given in radians

θ = 0.17 ° (π rad / 180 °) = 2.97 10⁻³ rad

d sin θ = m λ

λ = d sin θ / m

λ = 0.2 10⁻³ sin (2.97 10⁻³) / 2

λ = 3.119 10⁻⁷ m

8 0

2 years ago