How much force is needed to keep a 25 N rock moving with constant speed across the level, frictionless ice?

A boat is traveling upstream in the positive direction of the x axis at 10 km/h with respect to the water of a river. The water

sasho [114]

Answer:

The velocity of the boat with respect to the ground is 3 km/h

Explanation:

The speed of an object is different depending on the reference system you use. This is called relative speed.

A boat travels upstream, this means that it moves in the opposite direction to the river current.

A boat travels upstream, this means that it moves in the opposite direction to the river current. Then, if the boat moves in the positive direction of the x axis at 10 km / h with respect to the water of a river, the water flows in the negative direction of the x axis at 7 km / h with respect to the ground.

This causes the speed of the boat relative to the ground to be calculated as follows:

VbG = Vbw - VwG

where VbG is the speed of the boat relative to the ground, Vbw is the speed of the boat relative to the water of the river and VwG is the speed of the water relative to the ground.

So: VbG=10 km/h – 7 km/h

VbG= 3 km/h

The direction of this velocity is in the positive x-direction.

4 0

3 years ago

A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long sol

bearhunter [10]

Answer:

N= 3

Explanation:

For this exercise we must use Faraday's law

E = - dФ / dt

Ф = B . A = B Acos θ

tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives

E = - A cos θ (B - B₀) / t

The angle enters the magnetic field and the normal to the area is zero

cos 0 = 1

A = π r²

In the length of the wire there are N turns each with a length L₀ = 2π r

L = N (2π r)

r = L / 2π N

we substitute

A = L² / (4π N²)

The magnetic field produced by a solenoid is

B = μ₀ N/L I

for which

B₀ = μ₀ N/L I

The final field is zero, because the current is zero

B = 0

We substitute

E = - (L² / 4π N²) (0 - μ₀ N/L I) / t

E = μ₀ L I / (4π N t)

N = μ₀ L I / (4π t E)

The electromotive force is E = 0.80 mV = 0.8 10⁻³ V

let's calculate

N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]

N = 320 10⁻⁷ / 9.6 10⁻⁶

N = 33.3 10⁻¹

N= 3

7 0

2 years ago

If izzy mass is 0.3kg he applide 657.9n force what will be the accelration

Sholpan [36]

Answer:

The acceleration of the body, a = 2193 m/s²

Explanation:

Given,

The mass of the body, m = 0.3 kg

The force acting on the body, F = 657.9 N

The force acting on an object is proportional to the product of mass and acceleration of the body.

F = m x a

Therefore, the acceleration of the body is

a = F / m

= 657.9 N / 0.3 kg

= 2193 m/s²

Hence, the acceleration of the body, a = 2193 m/s²

4 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago