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3 years ago

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:

Explanation:

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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

$V_{T}$ = 1.77V

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