Ask question

Ask question

All categories

2 years ago

9

A car drives at a velocity of 6.8 m/s for 250 seconds. How far did it travel?

1 answer:

Leno4ka [110]2 years ago

8 0

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Here's the solution ~

As we know, Displacement =

$\qquad \sf \dashrightarrow \: velocity \times time$

$\qquad \sf \dashrightarrow \: 6.8 \times 250$

$\qquad \sf \dashrightarrow \: 1700 \: m$

<h3>OR </h3>

$\qquad \sf \dashrightarrow \: 1.7 \: km$

The car travelled 1700 meters ( 1.7 km )

You might be interested in

Is diet Pepsi heterogeneous or is it homogeneous

Lana71 [14]

Heterogeneous because it has less amount of sugar than the regular Pepsi has

6 0

3 years ago

Read 2 more answers

A solar cell, 3.0 cm square, has an output of 350 mA at 0.80 V when exposed to full sunlight. A solar panel that delivers close

Vesna [10]

Answer:

You will need 450 cells (3 cm each) to meet the voltage/current requirement.

The panel must be 3 cells in one side, by 150 cell in another side. 1350 cm^2 or 0.135 m^2. They must be connected 3 in row in parallel (to add current), then each of the former group must be connected in series to meet the voltage, so it would be 150 rows of connected in series.

The panel can be optimized using a voltage inverter, to convert current to voltage. In this way, less cells can be used achieving the same output specs.

Explanation:

To meet the voltage:

120 [v] required voltage

0.8 [v] voltage of each cell

$\frac{120}{0.8} =150[v]\\$

So we need 150 cells in series for the voltage.

To meet the current

1.0 [A] Required current

350[mA]=0.35[A] cell current

1/0.35=3 cell So we need 3 cells in parallel to add the currents and meet the requirement.

See the attached figure

8 0

3 years ago

Which has more heat, a home oven at 500 degrees Celsius, or a one ton commercial oven at 500 degrees celsius?

alisha [4.7K]

A commercial oven because it is more durable plus it’s more professional so they use more heat then we do at home

8 0

3 years ago

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

Why do you think spacecraft that carry instruments to collect data about objects in space are called probes?

yarga [219]

Answer:

my mind cant comprehend a shing word u just said

Explanation:

3 0

3 years ago

Read 2 more answers