HELP ASAPP

What is the approximate amount of thrust you need to apply to the lander to keep it's velocity roughly constant

earnstyle [38]

The approximate amount of thrust(force) you need to apply to the lander to

keep its velocity roughly constant is zero.

<h3>What is Newton's second law of motion?</h3>

Newton's second law of motion states that the acceleration the force acting

on the object is directly proportional to its rate of change of momentum.

F = m a

If the object is moving with uniform velocity, it simply means that the

acceleration is zero, and the corresponding force will also be zero.

Read more about Constant velocity here brainly.com/question/3052539

4 0

2 years ago

Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the

Natasha_Volkova [10]

The force per unit length between the two wires is $6.0\cdot 10^{-5} N/m$

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}$

where

$\mu_0 = 4\pi \cdot 10^{-7} Tm/A$ is the vacuum permeability

$I_1, I_2$ are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

$I_1 = 1.75 A$

$I_2 = 3.45 A$

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

$\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m$

Learn more about current and magnetic fields:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

4 0

3 years ago

The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from

grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Where

$\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux,

And $\phi=BA$

$\epsilon=-NA\dfrac{dB}{dt}$

$\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}$

$\epsilon=28.32\ V$

Using Ohm's law, $\epsilon=I\times R$

Induced current, $I=\dfrac{\epsilon}{R}$

$I=\dfrac{28.32}{1.5}$

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0

2 years ago

How to solve 2.14 using calculus

trasher [3.6K]

Answer:

The acceleration is $6.42\frac{m}{s^2}$

Explanation:

Given the velocity function:

$v=0.86\frac{m}{s^3}t^2$

you can obtain the instantaneous acceleration "a" as its first derivative:

$a=\dot{v}=2\cdot0.86\frac{m}{s^3}\cdot t=1.72\frac{m}{s^3}t$

To determine the value of "a" when the velocity was 12m/s, you need to figure out the value for "t" when this happens. At what time t is the velocity 12m/s?

$12.0\frac{m}{s}=0.86\frac{m}{s^3}t^2\implies t=3.74s$

This value of t is less than the 5 seconds mentioned in the text - so that is a good sign that the formula is valid for this value. And so you can use t=3.47s in the derivative (acceleration) above:

$a=1.72t=1.72\frac{m}{s^3}\cdot 3.74s = 6.42\frac{m}{s^2}$

3 0

3 years ago

Using the equation E = mc2, calculate how many joules of energy would be produced by converting 6.8 × 10-6 kg of matter into ene

mixer [17]

Answer:

Explanation:

Using the formula : E = mc²

Where m = mass = 6.8 × 10^-6 kg

c = speed of light = 3 ×10^8 m/s

The amount of joules of energy that would be produced equals :

Plugging in our values :

E = mc²

E = (6.8 × 10^-6) kg × (3 ×10^8)²m/s

E = (6.8 × 10^-6) kg × 9 × 10^16 m/s

E = 61.2 × 10^(-6 + 16)

E = 61.2 × 10^10 J

6 0

2 years ago