What is one way to lower gravitational potential energy?

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago

If the timber weighs 670 N, calculate its angle of inclination when the water surface is 2.1 m above the pivot. Above what depth

Ymorist [56]

Answer:

Q = 40.1 degrees

Explanation:

Given:

- The weight of the timber W = 670 N

- Water surface level from pivot y = 2.1 m

- The specific density of water Y = 9810 N / m^3

- Dimension of timber = (0.15 x 0.15 x 0.0036) m

Find:

- The angle of inclination Q that the timber makes with the horizontal.

Solution:

- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:

F_b = Y * V_timber

F_b = 9810*0.15*0.15*x

F_b = 226.7*x N

- Take static equilibrium conditions for the timber, and take moments about the pivot:

(M)_p = 0

W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0

- Plug values in:

670*0.5*3.6 - x^2 * 0.5*226.7 = 0

x^2 = 1206 / 113.35

x = 3.26 m

- Now use the value of x and vertical height y to compute the angle of inclination to be:

sin(Q) = y / x

sin(Q) = 2.1 / 3.26

Q = sin^-1 (0.6441718)

Q = 40.1 degrees

5 0

3 years ago

Can the velocity of a body revese the direction when acceleration is constant?

TEA [102]

Answer:

Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

7 0

3 years ago

Achilles and the tortoise are having a race. The tortoise can run 1 mile (or whatever the Hellenic equivalent of this would be)

fenix001 [56]

Answer:

Surely Achilles will catch the Tortoise, in 400 seconds

Explanation:

The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).

To calculate the time, we use kinematic expression for constant speed:

$x_{final}=x_{initial}+t_{tor}v_{tor}=1+t_{tor}\\x_{final}=x_{initial}+t_{ach}v_{ach}=10t_{ach}$

The moment that Achilles catch the tortoise is found by setting the same final position for both (and same time as well, since both start at the same time):

$1+t=10t\\t=1/9 hour=0.11 hours$

7 0

3 years ago