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2 years ago

Please answer please please Physics

Physics

1 answer:

zmey [24]2 years ago

6 0

Answer:

about -1.26 m/s²

Explanation:

The acceleration can be found from the formula ...

a = (v2² -v1²)/(2d)

Here, we have the initial velocity v1=23 m/s, the final velocity v2=0, and teh distance d=210 m. Using the formula, we find the acceleration to be ...

a = (0 -23²)/2(210) ≈ -1.2595 . . . m/s²

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after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

4 0

2 years ago

How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?

Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

$d = \frac{ \:Δv}{e}$

$= \frac{1.5v}{100v/m}$

$= 1.5 \times 10 {}^{ - 2} m$

learn more about potential difference from here: brainly.com/question/28166044

#SPJ4

6 0

4 months ago

A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl

Sati [7]

Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

6 0

3 years ago

Which explicit definition matches this sequence? an={10,if n=1an−1−4,if n>1 A. an = 10 + 4(n – 1) B. an = 10 – 4(n – 1) C. an

Step2247 [10]

Answer:

Añ=10+4(n-1)

Explanation:

3 0

2 years ago

What is the purpose of the report

AfilCa [17]

Answer:

The purpose of report : Reports communicate information which has been compiled as a result of research and analysis of data and of issues .

5 0

2 years ago

Read 2 more answers

Please answer please please Physics ​

Please answer please please Physics