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3 years ago

Atomic Nucleus: The atomic number of an atom is equal to the number of what particles in the nucleus?:

Physics

1 answer:

bulgar [2K]3 years ago

5 0

Answer:

B) protons

Explanation:

The total number of protons in an atom represents the atomic number of that element. An element is generally expressed as

$^N_Z{A}$

Where,

A is the element

N is the Mass number i.e., the number of neutrons and protons in an atom.

Z is the atomic number.

The elements are arranged according to their atomic number in the modern periodic table.

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A 1,680 kg satellite is in a circular orbit around Earth with a tangential speed of 6,578 m/s. What is the height of the satelli

balu736 [363]

Answer:

Recall that Earth’s radius is 6.38 × 106 m and Earth’s mass is 5.97 × 1024 kg.

Explanation:

8 0

3 years ago

Read 2 more answers

A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo

DIA [1.3K]

Answer:

$74^{\circ} C$

Explanation:

We are given that

Mass of glass, $m=300 g$

$T_1=23^{\circ}$

Volume, $V=236cm^3$

Mass of water= $density\times volume=1\times 236=236 g$

Density of water= $1g/cm^3$

Temperature of hot water, $T=87^{\circ}$

Specific heat of glass, $C_g=0.2cal/g^{\circ}C$

Specific heat of water, $C_w=1 cal/g^{\circ}C$

$Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)$

$Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)$

$Q_{glass}+Q_{water}=0$

$300\times 0.2(T_f-23)+236\times 1(T_f-87)$

$60T_f-1380+236T_f-20532=0$

$296T_f=20532+1380=21912$

$T_f=\frac{21912}{296}=74^{\circ} C$

5 0

3 years ago

A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bo

RoseWind [281]

Answer:

(a): a = 0.4m/s²

(b): α = 8 radians/s²

Explanation:

First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):

a= (Vf-Vi)/t = (2m/s)/t

a: linear acceleration.

Vf: speed at the end of the ramp.

Vi: speed at the beginning of the ramp (zero).

d= (1/2)×a×t² = 5m

d: distance of the ramp (5m).

We replace the first equation in the second to determine the travel time on the ramp:

d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s

And the linear acceleration will be:

a = (2m/s)/5s = 0.4m/s²

Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:

perimeter = π×diameter = π×0.1m = 0.3142m

To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:

α = (0.4m/s²)/(0.05m) = 8 radians/s²

α: angular aceleration.

3 0

3 years ago

a soccer player kicks a ball. According to Newton's third law of motion, the player's foot exerts a force on the ball. The ball

Leto [7]

Answer:

The forces are exerted on different objects so they are not balanced forces.

Explanation:

8 0

3 years ago

on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak

sineoko [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>

That's about 2.9 G's . Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't. Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

Force on the airplane = 504,000 Newtons

That's about 113,000 pounds ! Maybe the part of the airplane that the catapult pushes on can't handle any more force than that. Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds. Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes ! Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

8 0

3 years ago