What are the basic properties of matter

What is the kinetic energy of a 0.135 kg baseball thrown at 40 m/s

Ronch [10]

KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J

4 0

3 years ago

What are dot diagrams in Physical Science?

s344n2d4d5 [400]

Answer:

If you mean Lewis dot diagrams, aka electron-dot diagrams, then these are diagrams that show the bonding between atoms of a molecule, and the lone pairs of electrons that may exist in the molecule.

Explanation:

6 0

3 years ago

Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a spee

abruzzese [7]

Answer:

Explanation:

Mass of proton :

$m_P=1.67\times 10^-^2^7\:kg\\$

Speed of Proton:

$v_P=4.85\times 10^6$

Linear Momentum of a particle having mass (m) and velocity (v) :

$-> p =m->v\:\:\: (1)$

Magnitude of momentum :

$p=mv\:\:\: (2)$

Frome equation (2), magnitude of linear momentum of the proton :

$p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1$

7 0

3 years ago

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

matrenka [14]

(a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
 $F=qE$
As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
 $F=qvB \sin \theta$
where $\theta$ is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
 $F=qE$
as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

(d) does a magnetic field do so?
Yes. As we said in point b, the magnetic force is
$F=qvB \sin \theta$
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
 $F=qE$
So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
 $F=qvB \sin \theta$
is different from zero because v is different from zero.

6 0

3 years ago

A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb

marin [14]

Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

$n_1 =\dfrac{46 \times 1}{0.0821\times 298}$

n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

$n_2 =\dfrac{12 \times 1}{0.0821\times 298}$

n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

$P_1=\dfrac{n_1 R T}{V}$

$P_1=\dfrac{1.89\times 0.0821\times 298}{5}$

P₁ = 9.25 atm

Partial pressure of oxygen

$P_2=\dfrac{n_2 R T}{V}$

$P_2=\dfrac{0.49\times 0.0821\times 298}{5}$

P₂ = 2.39 atm

total pressure

P = P₁ + P₂

P = 9.25 + 2.39

P = 11.64 atm

8 0

3 years ago