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Ainat [17]
3 years ago
10

An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?

Physics
2 answers:
Dimas [21]3 years ago
8 0

Answer:

35

Explanation:

dfddffffffffffffdddfr

dsp733 years ago
3 0

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

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Consider a large 1.54 V carbon-zinc dry cell used in a physics lab to supply 2.15 A to a circuit. The internal resistance of the
andrew-mc [135]
<h2>Answer:</h2>

1.77V

<h2>Explanation:</h2>

The electromotive force voltage (E) in a cell, is related to the lost voltage (V_{L}) and the terminal voltage (V_{T}) as follows;

E = V_{T} - V_{L}

Where;

The lost voltage (V_{L}) is the product of the internal resistance (r) of the cell and current (I) in the cell. i.e

V_{L} =  I x r

<em>Substitute </em>V_{L}<em> =  I x r into equation (i) as follows;</em>

E = V_{T} - (I x r)           ----------------------(ii)

<em>According to the question;</em>

E = 1.54V

I = 2.15A

r = 0.105Ω

<em>Substitute these values into equation(ii) as follows;</em>

1.54 = V_{T} - (2.15 x 0.105)

1.54 = V_{T} - (0.22575)

1.54 = V_{T} - 0.22575

<em>Solve for </em>V_{T}<em>;</em>

V_{T} = 1.54 + 0.22575

V_{T} = 1.54 + 0.22575

V_{T} = 1.77V

Therefore, the terminal voltage of the cell is 1.77V

8 0
3 years ago
Bill and amy want to ride their bikes to school which is 14.4 kilometers away. It takes Amy 49 minutes to get to school and bill
Drupady [299]

3.4m/s

Explanation:

Given parameters:

Distance to school  = 14.4km

Time taken by Amy = 49min

Time taken by bill = 20min after Amy = 20+49 = 69min

Unknown parameters:

How much faster is Amy's average speed = ?

Solution:

Average speed is the rate of change of total distance with total time taken.

 Average speed = \frac{total distance }{total time taken}

convert units to meters and seconds

      1000m = 1km

       60s = 1min

Distance to school  = 14.4 x 1000 = 14400m

Time taken by Amy = 49 x 60 = 2940s

Time taken by Bill = 69 x 60 = 4140s

Average speed of Amy = \frac{14400}{2940}  = 4.9m/s

Average speed of Bill = \frac{4140}{2940}  = 1.4m/s

Differences in speed = 4.9 - 1.5 = 3.4m/s

Amy was 3.4m/s faster than Bill

learn more:

Average speed brainly.com/question/8893949

#learnwithBrainly

5 0
3 years ago
Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p
ira [324]

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

8 0
3 years ago
Approximately what percentage of incoming solar radiation is absorbed by the oceans and continents
AleksandrR [38]
90 percent a day to keep things running smoothly
8 0
3 years ago
The potential difference between two equipotential lines 5.0 mm apart has a value of 1.2 V, calculate the electric field value.
jonny [76]

Answer:

The electric field value is 240 N/C

Explanation:

Given that,

Distance = 5.0 mm

Potential difference = 1.2 V

We need to calculate the electric field value

Using formula of potential difference

\Delta V=Ed

E=\dfrac{\Delta V}{d}

Where, E = electric field

V = potential difference

d = distance

Put the value into the formula

E=\dfrac{1.2}{5.0\times10^{-3}}

E= 240\ N/C

Hence, The electric field value is 240 N/C

6 0
3 years ago
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