A 0.6 m length of current carrying wire is placed in a uniform magnetic field oriented perpendicularly to the wire. The current is 4 A and the magnitude of the magnetic field is 10 T. What is the force on the wire?
a. 0 N.
b. 2.4 N.
c. 6 N.
d. 24 N.
e. 30 N.

Answer:

The final velocity is 20 m/s.

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

[tex]v_f=v_o+at[/tex]

The provided data is: vo=10 m/s, [tex]a=5\ m/s^2[/tex], t=2 s. The final velocity is:

[tex]v_f=10~m/s+5\ m/s^2\cdot 2\ s[/tex]

[tex]v_f=20\ m/s[/tex]

The final velocity is 20 m/s.

Answer:

options B,C,D are true

Explanation:

Answer:

s = 3.26 m

Explanation:

Given that,

Water leaves at a speed of 8 m/s

We need to find the maximum height that steelhead can jump. Let it can jump to a height of h.

At maximum height, final speed is equal to 0. We can use third equation of motion to find the maximum height.

[tex]v^2-u^2=2as[/tex]

a = -g

[tex]-u^2=-2gs\\\\s=\dfrac{u^2}{2g}\\\\s=\dfrac{(8)^2}{2\times 9.8}\\\\=3.26\ m[/tex]

Hence, the maximum height is 3.26 m.

Answer:

STOP CHETING

Answer:

C is the answer to your question

Explanation:

Have a great day!!! :)

Electrical power can be defined as the product of Voltage and current. Thus, P = IV. Hence, Option D is the correct answer.

Thus, the formula for electric power,

P = IV watt

Where,

P - Power (Watt)

I - Current (Ampere)

V - Voltage (volts)

Electrical power can be defined as the product of Voltage and current. Thus, P = IV. Hence, Option D is the correct answer.

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Answer:

A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.

Explanation:

A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.

Answer:

Explanation:

for rolling motion down the plane acceleration is given by the following expression

a = g sinθ / (1 + k² / R²)

here k is radius of gyration and R is radius of the object rolling down .

for cylinder I = 1/2 m R²

so k² = R² / 2

k² / R² = 1/2

a = g sinθ /( 1 + 1 / 2 )

= 2 / 3 x  g sinθ

v = √ 2 a s

= √ (2 x  2 / 3 x  g sinθ s )

= √ (4  / 3 x  g h  )

= √ (4  / 3 x  g x .5  )

= √ 2g / 3

for sphere  I = 2/5  m R²

so k² = 2/5 R²

k² / R² = 2 / 5  

a = g sinθ / (1 + 2 / 5)  

= 5 / 7  x  g sinθ

v = √ 2 a s

= √ (2 x  5 / 7  x  g sinθ s )

= √ (10/7  x  g h  )

Given

√ (10/7  x  g h  ) = √ 2g / 3

10/7  x  g h  = 2g / 3

h = 14 / 30 m

= .47 m .

Explanation:It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

Answer:

1717 kg*m/s

Explanation:

found it put it in on acellus and it was right

Answer:

I think it's most likely true.

Explanation:

any organism has the properties of a living thing, which includes cells, whether it has one cell or many

Answer:

False

Explanation:

An organism is a living thing that is a single-celled life form

Answer:

44 grams- 55 grams through the whole day. Probably about 14.6 grams per meal.

Explanation:

Answer:

2 thumbs (ISSA Guide)

Explanation:

Answer:

light refraction

Explanation:

Answer:

[tex]0.566\; \rm N[/tex] (assuming that while the block is moving, the friction on the block is constant.)

Explanation:

The mechanical energy of a system is the sum of its:

Friction does work on the block as the block moves up the ramp. The amount of energy that the block-string system has lost would be equal in size to the work that friction has done on the block. The size of the friction on the block could thus be computed.

Before this block was released, the block-spring system has no kinetic energy because there was no movement. Assume that the system has no gravitational potential energy at that moment, either. The only type of mechanical energy in this system at that moment would be elastic potential energy: [tex]8\; \rm J[/tex] according to the question.

The question states that the ramp is [tex]53.0^\circ[/tex] above horizontal. Therefore, after the block has traveled [tex]3.00\; \rm m[/tex] (along the ramp,) the height of this block would have increased by [tex]\Delta h = 3.00\; \rm \sin\left(53^\circ\right) \approx 2.39591\; \rm m[/tex]. Calculate the corresponding gain in gravitational potential energy:

[tex]\begin{aligned}& m \cdot g \cdot \Delta h\\ &\approx 0.200\; \rm kg& \\\ &\quad\times 9.81\; \rm N \cdot kg^{-1} \\ &\quad \times 2.39591\; \rm m \\ &\approx 4.70077\; \rm J\end{aligned}[/tex].

On the other hand, the question states that the speed of the block ([tex]m = 0.200\; \rm kg[/tex]) at that moment is [tex]v = 4.00\; \rm m \cdot s^{-1}[/tex]. Calculate the corresponding kinetic energy:

[tex]\begin{aligned}& \frac{1}{2}\, m \cdot v^{2} = 1.6\; \rm J \end{aligned}[/tex].

Because the block was already released, there should be no elastic potential energy in the spring.

Hence, the mechanical of the block-spring system would be approximately [tex]4.70077\; \rm J+ 1.6\; \rm J \approx 6.30077\; \rm J[/tex].

Approximate the amount of mechanical energy that is lost:

[tex]8\; \rm J - 6.30077\; \rm J \approx 1.69923\; \rm J[/tex].

In other words, when applied over [tex]3\; \rm m[/tex], the friction on this block would do approximately [tex]1.69923\; \rm J[/tex] of work. Approximate the size of that friction:

[tex]\begin{aligned}F &= \frac{W}{s} \\ &\approx \frac{1.69923\; \rm J}{3.00\; \rm m}\approx 0.566\; \rm N\end{aligned}[/tex].

Answer:

20 m/s^2

Explanation:

Use the formula Vf=Vi+at

Vf=100

Vi=0

a=?

t=5

Then solve for (a)

Answer:B

Explanation: i took the test and i got it right

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 60 × 2

We have the final answer as

Hope this helps you

Answer:

It’s a

Explanation:

Don’t actually put that i needed the points mb

Answer:

jack has done more work pulling more weight and Jill has more power.

Explanation:

Answer:

Explanation:

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.1 m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2403.07 x 10⁻⁶ m

= 2.403 x 10⁻³ m

= 2.403 mm .

For λ = 655 nm

position = 2 λ D / d

λ = 655 nm , D = 1.1 m

d = .65 x 10⁻³

position 2 = 2 x 655 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2216.91 x 10⁻⁶ m

= 2.217 x 10⁻³ m

= 2.217 mm .

Difference between their position

= 2.403 - 2.217 = .186 mm .

Answer:

Approximately 1.02 seconds

Explanation:

Use the final velocity (vf) formula for a uniformly accelerated movement under "g" (acceleration of gravity):

[tex]v_f=v_i+g\,*\,t[/tex]

in our case:

[tex]10=0+9.8\,*\,t\\t=10/9.8\\t\approx 1.02\,\,sec[/tex]

Answer:

F = 288 [N]

Explanation:

To solve this problem we must use the following equation of kinematics and find the value of acceleration.

[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]

where:

Vf =  final velocity = 6 [m/s]

Vo =  initial velocity = 0 (starting from rest)

a = acceleration [m/s²]

x = distance = 5 [m]

Now replacing, we have:

[tex](6)^{2}=0+(2*a*5)\\36=10*a\\a = 3.6 [m/s^{2}][/tex]

Since we already have the value of acceleration, we can use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

[tex]F =80*3.6\\F = 288 [N][/tex]

Answer:

Explanation:

The speed of the horse can be found by using the formula

[tex]v = \frac{p}{m} \\ [/tex]

p is the momentum

m is the mass

From the question we have

[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]

We have the final answer as

Hope this helps you

Answer:

[tex]-1.398\ \text{m/s}^2[/tex]

Decelerating or slowing down

Explanation:

F = Force = 50 N

[tex]\theta[/tex] = Angle force is being applied = [tex]42^{\circ}[/tex]

[tex]\mu[/tex] = Coefficient of friction = 0.64

m = Mass of wagon = 12 kg

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Normal force is given by

[tex]N=mg-F\sin\theta[/tex]

Frictional force is given by

[tex]f=\mu N\\\Rightarrow f=\mu (mg-F\sin\theta)[/tex]

The force balance is given by

[tex]F\cos\theta-f=ma\\\Rightarrow \dfrac{F\cos\theta-\mu (mg-F\sin\theta)}{m}=a\\\Rightarrow a=\dfrac{50\times \cos42^{\circ}-0.64(12\times 9.81-50\times\sin42^{\circ})}{12}\\\Rightarrow a=-1.398\ \text{m/s}^2[/tex]

The acceleration of the wagon is [tex]-1.398\ \text{m/s}^2[/tex]. The negative sign indicates that the wagon is decelerating or slowing down.

The acceleration of the wagon is  [tex]-1.398 \;\rm m/s^{2}[/tex].

Given data:

The magnitude of pulling force is, F = 50 N.

The angle of inclination is, [tex]\theta = 42^{\circ}[/tex].

The mass of wagon wheel is, m = 12 kg.

Coefficient of friction between wagon and grass is, [tex]\mu =0.64[/tex].

The given problem is based on the concept of frictional force. The standard expression for the frictional force is,

[tex]f= \mu \times N[/tex]

Here, N is the normal force and its value is,

[tex]N=mg-Fsin \theta[/tex]

And the net force acting on wagon is,

[tex]F' = Fcos\theta -f\\\\ma = Fcos\theta -(\mu(mg-Fsin \theta))\\\\a = \dfrac{Fcos\theta -(\mu(mg-Fsin \theta))}{m}[/tex]

Here, a is the acceleration of wagon.

Solving as,

[tex]a = \dfrac{50 \times cos42 -(0.64(12 \times 9.8-(50 \times sin42)))}{12}\\\\a=-1.398 \;\rm m/s^{2}[/tex]

Thus, we can conclude that the acceleration of the wagon is  [tex]-1.398 \;\rm m/s^{2}[/tex].

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Answer:

a)t  = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s   ;    V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Explanation:

According to the problem statement

V₀ = V₀ₓ    and  V₀y = 0

And at the end of the movement t = ?  the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9    ⇒  t² = 2,04 s

t  = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ        V₀y = 0     and  V₀ₓ = Vₓ     ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then  15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s    since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf =  0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s

Answer:

The sun is the ultimate source of energy for all food chains. Through the process of photosynthesis, plants use light energy from the sun to make food energy. Energy flows, or is transferred through the system as one organism consumes another.

Answer:

chemical symbol: Th

atomic number:90

protrons :90

neutrons:142

group#:4

period#: 9

Explanation:

you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons

Answer:

a. The bowling ball would have more kinetic energy because of its greater mass.

b. Potential energy

Explanation

a. Bowling ball has higher mass, self explanatory.

b. A high diver has lots of stored energy when they are on the diving platform. When they dive this stored energy helps make the splash when they hit the water. Stored energy is also called potential energy.