A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.

Answers

Answer 1

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:


Related Questions


2. Using a giant screw, a crew does 650 J of work to drill a hole into a rock.
The screw does 65 J of work. What is the efficiency of the screw? Show your
work. Hellpppp

Answers

yeah un jsjsjsjjsos isnsisoowam

Answer:

42,250

Explanation:

It goes inside=

Displacemt

It does work=

Work done

To find efficiency of jule we do=

Dicplacement × Work done

650 × 65

42,250

Please mark me as a brainlist

Derase
An electric heater Consumes 1.8 MJ When connected to a 250V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply

Answers

Answer:

a. Power = 1000 Watts or 1 Kilowatts.

b. Current = 4 Amperes.

Explanation:

Given the following data;

Energy consumed = 1.8MJ = 1.8 × 10^6 = 1800000 Joules

Voltage = 250V

Time = 30 minutes to seconds = 30 * 60 = 1800 seconds

To find the power rating;

Power = energy/time

Substituting into the equation, we have;

Power = 1800000/1800

Power = 1000 Watts or 1 Kilowatts.

b. To find the current taken from the supply;

Power = current * voltage

1000 = current * 250

Current = 1000/250

Current = 4 Amperes.

Which is an example of kinetic energy?

A. The energy stored in
ethanol

B. A ball sitting at the top of a ramp

C. A compressed spring

D. A hockey puck sliding across ice

Answers

D . A hockey puck sliding across ice

D. A hockey puck sliding across ice

Help plsssssssssss I write it 100 time no one answer

Answers

Answer:

1.93×10²⁸ s

Explanation:

From the question given above, the following data were obtained:

Number of electron (e) = 2×10²⁴

Current (I) = 10 A

Time (t) =?

Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:

1 e = 96500 C

Therefore,

2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e

2×10²⁴ e = 1.93×10²⁹ C

Thus, 1.93×10²⁹ C of electricity is passing through the point.

Finally, we shall determine the time. This can be obtained as follow:

Current (I) = 10 A

Quantity of electricity = 1.93×10²⁹ C

Time (t) =?

Q = it

1.93×10²⁹ = 10 × t

Divide both side by 10

t = 1.93×10²⁹ / 10

t = 1.93×10²⁸ s

Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point

is 250 000 miles from the earth to the moon" is a qualitative
Observation
TRUE
Or false

Answers

Answer:

True

Explanation:

A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.​

Answers

Answer:

The speed of this particle is constantly [tex]c[/tex].

Explanation:

Position vector of this particle at time [tex]t[/tex]:

[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].

Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:

[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].

Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].

Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:

[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].

The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :

[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].

Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)

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