Should the resistance of an ammeter low or high ?give reason

Answer:

The value is [tex]T_t = 2.5659 \ s[/tex]    

Explanation:

From the we are told that  

         The initial speed of the object is  [tex]u = 8 \ m/s[/tex]

         The greatest height it reached is  [tex]h = 15 \ m[/tex]

Generally from kinematic equation we have that

      [tex]v^2 = u^2 + 2gH[/tex]

At maximum height v  =  0 m/s

So

      [tex]0^2 = 8^2 + 2 * - 9.8 * H[/tex]

=>    [tex]H = 3.27 \ m[/tex]

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

      [tex]s = h - H[/tex]

=>    [tex]s = 15 - 3.27[/tex]

=>    [tex]s = 11.73 \ m[/tex]

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

            [tex]v = u + (-g) t[/tex]

At maximum height v  =  0 m/s

           [tex]0 = 8 - 9.8t[/tex]

=>         [tex]t = 0.8163 \ s[/tex]

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       [tex]h = ut_1 + \frac{1}{2} gt_1^2[/tex]

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       [tex]15 = 0* t + 4.9 t^2[/tex]

=>      [tex]t_1 = 1.7496[/tex]

Generally the total time taken is mathematically represented as

          [tex]T_t = t + t_1[/tex]

=>        [tex]T_t = 0.8163 + 1.7496[/tex]

=>        [tex]T_t = 2.5659 \ s[/tex]            

Answer:

forever

Explanation:

Answer:

idgaf

Explanation:

lol

Answer:

[tex]\eta_{turbine} = 0.777 = 77.7\%[/tex]

[tex]\eta_{combined} = 0.728 = 72.8\%[/tex]

Explanation:

First we calculate the power input to the turbine. The input power will be equal to the potential energy of water per unit time:

Input Power = [tex]P_{in} = \frac{Work}{Time} = \frac{Potential\ Energy\ of\ Water}{t} \\P_{in} = \frac{(mass)(g)(height)}{Time} = (mass flow rate)(g)(height)\\\\P_{in} = (1500\ kg/s)(9.81\ m/s^2)(70\ m)\\P_{in} = 1.03\ x\ 10^6\ W = 1030\ KW[/tex]

Now, for turbine efficiency:

[tex]\eta_{turbine} = \frac{Mechanical\ Power\ Out}{P_{in}}\\\\\eta_{turbine} = \frac{800\ KW}{1030\ KW}\\\\\eta_{turbine} = 0.777 = 77.7\%[/tex]

for generator efficiency:

[tex]\eta_{generator} = \frac{Power\ Generation}{Mechanical\ Power\ Out}\\\\ \eta_{generator} = \frac{750\ KW}{800\ KW}\\\\\eta_{turbine} = 0.9375 = 93.75\%[/tex]

Now, for combined efficiency:

[tex]\eta_{combined} = \eta_{turbine}\ \eta_{generator}\\\\\eta_{combined} = (0.777)(0.937)\\\eta_{combined} = 0.728 = 72.8\%[/tex]

Answer:

[tex]\huge\boxed{10\:watts}[/tex]

Explanation:

Power = Current × Voltage

P = I × V

P = 2 × 5

P = 10 watts

Answer:

a= g = - 9.81 m/s2.

The following equations will be helpful:

a = (vf - vo)/t d = vot + 1/2 at2 vf2 = vo2 + 2ad

When you substitute the specific acceleration due to gravity (g), the equations are as follows:

g = (vf - vo)/t d = vot + 1/2 gt2 vf2 = vo2 + 2gd

If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:

g = vf /t d = 1/2 gt2 vf2 = 2gd

The sign convention that we will use for direction is this: "down" is the negative direction. If you are given a velocity such as -5.0 m/s, we will assume that the direction of the velocity vector is down. Also if you are told that an object falls with a velocity of 5.0 m/s, you would substitute -5.0 m/s in your equations. The sign convention would also apply to the acceleration due to gravity as shown above. The direction of the acceleration vector is down (-9.81 m/s2) because the gravitational force causing the acceleration is directed downward.

hope this info helps you out!

Answer:

68.2 Quadrillion meters

Explanation:

A lightyear is the distance that light travels in one year.

Speed of light is [tex]3*10^8\ m/s[/tex]

So light covers 300,000,000 meters in one second.

One year has 31536000 seconds so , light covers

[tex]9.461*10^{15}\ meters\ in\ one\ year[/tex]

so 7.2 light years is

[tex]7.2*(9.461*10^{15})\\6.82*10^{16}[/tex]

so 7.2 light years is

6.82 x 10^(16) meters or

68.2 Quadrillion meters

Answer:

A. im not for sure hope it helps

Explanation:

Answer:

a. 3.68 m b. 0.61 m c. 3.07 m

Explanation:

a. What is the maximum height the ball reaches on the moon?

Using v² = u² - 2g'h where u = initial velocity of baseball = 12 m/s, v =  final velocity of baseball = 0 m/s (since it stops at maximum height), g' = acceleration due to gravity on moon = g/6 where g = acceleration due to gravity on earth = 9.8 m/s², so g' = 9.8 m/s²/6 = 1.63 m/s² and h = maximum height of ball on moon.

So, making h subject of the formula, we have

h = -(v² - u²)/2g'

substituting the values of the variables, we have

h = -((0 m/s)² - (12 m/s)²)/2(1.63 m/s²)

=  -(- 12 m²/s²)/2(1.63 m/s²)

= 6 m²/s²)/(1.63 m/s²)

= 3.68 m

b. What is the maximum height the ball reaches on the earth?

Using the same equation for the maximum height the baseball travels on the moon for that on the earth, the maximum height h' the baseball travels on the earth is given by  

h' = -(v² - u²)/2g where u = initial velocity of baseball = 12 m/s, v =  final velocity of baseball = 0 m/s (since it stops at maximum height),  g = acceleration due to gravity on earth = 9.8 m/s²and h' = maximum height of ball on moon.

substituting the values of the variables, we have

h = -((0 m/s)² - (12 m/s)²)/2(9.8 m/s²)

=  -(- 12 m²/s²)/2(9.8 m/s²)

= 6 m²/s²)/(9.8 m/s²)

= 0.61 m

c. What is the difference in the maximum height the ball reaches on the Moon compared to on Earth?

The difference in the maximum height the ball reaches on the Moon compared to on Earth is d = h - h' = 3.68 m - 0.61 m = 3.07 m

Answer:

i believe it's 7

Explanation:

Answer:

5 years worth of work (aka all of the homework i currently have)

Answer:

Explanation:

Given parameters:

Weight of object  = 49N

Force applied = 12N

Unknown:

Acceleration of object  = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

 Acceleration  = [tex]\frac{12}{49}[/tex]   = 0.25m/s²