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3 years ago

What do the spheres in the model represent

Physics

2 answers:

Reptile [31]3 years ago

5 0

Make another question with a picture

Sophie [7]3 years ago

4 0

Answer: no

Explanation: we need a picture

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_____ is not a hormone.

Arlecino [84]

Answer:

So, the correct answer is 'Melatonin'.

Explanation:

thank me later

4 0

3 years ago

A briefcase sits stationary in an elevator. The mass of the briefcase if 4.5 kg. The elevator then begins accelerating upwards a

Korvikt [17]

Answer:

D. 48.985 N

Explanation:

Newton's second law states that:

$\sum F = ma$

which means that the net force acting on an object is equal to the product between the object's mass and its acceleration.

The equation of the forces for the briefcase in the elevator therefore is given by:

$N-mg=ma$

where

N is the normal reaction exerted on the briefcase

(mg) is the weight of the briefcase, with

m = 4.5 kg being its mass

g = 9.8 m/s^2 is the acceleration of gravity

a = 1.10 m/s^2 is the acceleration

Here we chose upward as positive direction.

Solving for N, we find the normal force:

$N=mg+ma=m(g+a)=(4.5)(9.81+1.10)=49.095 N$

So the closest answer is

D. 48.985 N

3 0

3 years ago

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

$E = \dfrac{1}{2}mv^2$

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago

Take the following measured gauge (or gage) pressures and convert them to absolute pressures in both kPa and psi units for an am

mamaluj [8]

Answer:

Explanation:

Given

ambient Pressure =98.10 kPa

(a)gauge pressure 152 kPa

we know

Absolute pressure=gauge pressure+Vacuum Pressure

$P_{abs}$ =152+98.10=250.1 kPa or 36.27 psi

(b) $P_{gauge}$ =67.5 Torr or 8.99 kpa

as 1 Torr is 0.133 kPa

$P_{abs}$ =8.99+98.10=107.09 kPa or 15.53 psi

Thus absolute pressure=98.10-10=88.10 kPa or 12.77 psi

as 1 kPa is equal to 0.145 psi

(d) $P_{vaccum}$ =0.84 atm or 85.113 kPa

as 1 atm is equal to 101.325 kPa

$P_{abs}$ =98.10-85.11=12.99 kPa or 1.88 psi

6 0

3 years ago

A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, w

Annette [7]

Answer:

66 rpm

Explanation:

The period of oscillation is given by

$T=2\pi \sqrt{\frac {m}{k}}$

$\frac {k}{m}=\frac {4\pi^{2}}{T^{2}}$ where T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.

Also, net force is given by

Net force= $m\omega^{2} L$

$\omega=\sqrt{\frac {k\triangle L}{mL}}$ where $\triangle L$ is the elongation, L is original length, $\omega$ is the angular velocity

Substituting the equation of $\frac {k}{m}$ into the above we obtain

$\omega=\sqrt {\frac {4\pi^{2}\triangle L}{T^{2} L}}$

$\omega=\sqrt {4\pi^{2}\times 0.15L}{0.35^{2}\times L}}=6.952763\approx 6.95 rad/s$

$6.95\times\frac {60 s}{2\pi rad}\approx 66 rpm$

6 0

4 years ago