All categories

3 years ago

What is the acceleration of a proton moving with a speed of 7.0 m/s at right angles to a magnetic field of 1.7 t ?

1 answer:

8 0

We know that a charge moving in a magnetic field is subject to the force:
F = q · v · B

But we also know that:
F = m · a

Therefore, it must be:
m · a = q · v · B

And solving for a:
a = q · v · B / m

Recall that for a proton:
q = 1.6×10⁻¹⁹ C
m = 1.673×10⁻²⁷ kg

Now, you can find:
a = 1.6×10⁻¹⁹ · 7.0 · 1.7 / 1.673×10⁻²⁷
= 1.14×10⁹ m/s²

Hence, the acceleration of the proton is 1.14×10⁹ m/s².

You might be interested in

Which statement correctly compares sound and light waves

kolbaska11 [484]

Answer:

Light waves carry energy parallel to the motion of the wave, while sound waves carry energy perpendicular to it. Sound waves carry energy parallel to the motion of the wave, while light waves carry energy perpendicular to it.

Explanation:

8 0

3 years ago

g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn

krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

T² = ( $\frac{4\pi }{G M_s}$ ) r³ (1)

in this case the period of the season is

T₁ = 93 min (60 s / 1 min) = 5580 s

r₁ = 410 + 6370 = 6780 km

r₁ = 6.780 10⁶ m

for the satellite

T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

T² = K r³

K = T₁²/r₁³

K = $\frac{ 5580^2}{ (6.780 10^6)^2}$

K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

r³ = T² / K

r³ = 86400² / 9.99 10⁻¹⁴

r = ∛(7.4724 10²²)

r = 4.21 10⁷ m

this distance is from the center of the earth

7 0

3 years ago

When a gas is rapidly compressed (say, by pushing down a piston) its temperature increases. When a gas expands against a piston,

shusha [124]

Answer:

Explained in explanation

Explanation:

The first law of thermodynamics states that the change in internal energy of a system(ΔU) is equal to the sum of the net heat transfer into the system(Q) and the net work done on the system(W). In equation, this law is;

ΔU = Q + W

Now, when there's gas inside a container with a movable piston that's tightly fitting, we will assume that the piston can move up and down thereby compressing the gas or allowing the gas to expand against it.

Now these gas molecules inside the container possess kinetic energy. Thus, the internal energy(U) of the system is simply the sum of all the kinetic energies of the individual gas molecules present in the container.

Therefore, if the temperature(T) of the gas increases, then the speed and internal energy(U) of the gas molecules will also increase. In the same way, if the temperature of the gas decreases, the speed and internal energy of the gas molecules would also decrease.

Now, back to the question, when the piston is pushed down, it does work on the gas and the gas does negative work on the piston. Thus, the gas will be get compressed to a smaller space, and thereby making the gas molecules to hit the piston at a faster rate. Thus, there is a decrease in speed and as we saw earlier that when there is a decrease in speed, it means temperature has decreased.

Whereas, when the piston is moved up, the gas does positive work on the piston and the speed of the gas molecules will increase. Like I said earlier that increase in speed means increase in temperature.

4 0

3 years ago

When a wire 1.5 m long carries a 24-A current in the +x direction in a uniform external magnetic field, the magnetic force exert

VLD [36.1K]

Answer:

Explanation:

Let the magnetic field be

B = $B_xi +B_yj +B_z k$

For magnetic force , the expression is

F = L ( I x B )

= 1.5 ( 24i x $B_xi +B_yj +B_z k$ )

F = 36 B_y k - 36 B_z j

Given

F = 3 j + 2 k

Equating equal terms

we have

B_y = 2 / 36 , B_z = - 3 / 36

Now the direction of current is changed to y direction

so F = $1.5[ 24j \times ( B_xi +B_yj +B_z k )]$

Given

F = - 3 i - 2 k

Equating equal terms

we have

B_x = 2 / 36 , B_z = - 3 / 36

So B = 2/36 i + 2/36 j - 3/ 36 k

Magnitude of B

= 4.1 / 36 T

8 0

3 years ago

A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision w

patriot [66]

Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

4 0

3 years ago