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dedylja [7]
2 years ago
11

Question 8 (1 point)

Physics
1 answer:
sineoko [7]2 years ago
6 0

Answer:

A

Explanation:

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A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
2 years ago
The temperature on the moon is
Nuetrik [128]
B.

It can go from very hot to very cold, it depends on the area of the moon and where the sunlight hits.
4 0
3 years ago
Read 2 more answers
A 50 n force is acting on a lever 1.5 m from the fulcrum balances an object 1m from the fulcrum on the other arm. what is the we
Angelina_Jolie [31]

Answer:

A 100 N force acting on a lever 2 m from the fulcrum balances an object 0.5 m from the fulcrum on. ... What is the weight of the object(in newtons)? What is its mass (in kg)? ... mass at the one end and effort arm is the distance between pivot and effort applied at the other end.

Explanation:

hpoe this helps you.

4 0
2 years ago
An electron is projected vertically upward with a speed of 1.70 ✕ 106 m/s into a uniform magnetic field of 0.320 t that is direc
Serggg [28]
The electron's path in the magnetic field is a straight line when viewed from above.

In fact, the electron initially moves upward, while the magnetic field is directed horizontally. The electron experiences a force due to the magnetic field (the Lorentz force), whose direction is given by the right-hand rule:
- index finger --> initial direction of the electron (upward)
- middle finger --> direction of the magnetic field (horizontally, away from the observer)
- opposite direction to the thumb* --> direction of the force (horizontally, but perpendicular to the magnetic field, to the right)

This means that the Lorentz force makes the electron moving perpendicular to the magnetic field in the horizontal plane, and since the direction of the field is not changing, this force does not change its direction, so the electron moves in the same direction of the force in the horizontal plane (to the right), therefore following a straight line.

* the direction should be reversed because the charge is negative.
7 0
3 years ago
Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a sta
yuradex [85]

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

7 0
3 years ago
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