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3 years ago

What is the electric field strength 7.50 x 10E-1 meters from a 2.8 x 10 E-3 C charged object?

Physics

1 answer:

Nadusha1986 [10]3 years ago

5 0

Answer:hi

Explanation:

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The first asteroid to be discovered is Ceres. It is the largest and most massive asteroid in our solar system’s asteroid belt, h

stira [4]

Answer:

$4.81*10^{29}J$

Explanation:

Since the formula for kinetic energy of an object is:

$E_k = \frac{mv^2}{2}$

Where m is the mass of the object and v is the speed. We can substitute m = $3*10^{21}$ kg and v = 17900m/s:

$E_k = \frac{3*10^{21} * (17900)^2}{2} = 4.81*10^{29}J$

3 0

2 years ago

9. A 12 v battery is connected to four 5 ohm light bulbs. What is the equivalent

vlabodo [156]

20 ohms 5 ohms
12volts

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2 years ago

At a height of ten meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the la

konstantin123 [22]

Answer:

Explanation:

Velocity of sound in air at 20 degree = 343 m/s

Velocity of sound in water at 20 degree = 1470 m/s

Time taken in to and fro movement in air

=( 2 x 10) / 343 = 0.0583 s

Rest of the time is

.171 - .0583 = .1127 s

This time is taken to cover distance in water. If d be the depth of lake

2d / velocity = time taken

2 d / 1470 = .1127

d = 82.83 m

5 0

2 years ago

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before

Darina [25.2K]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

$\\$

☯ As we know that,

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

$\\$

$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

$\\$

$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

$\\$

$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

$\\$

$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

$\\$

$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

$\\$

$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

$\\$

$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

4 0

2 years ago

A biker is pedaling at a constant speed of 36 km/h. During the last 10 s of the race, he increases his speed with a constant acc

adell [148]

Answer:

54 km/h

Explanation:

given,

speed of the biker = 36 Km/h

time = 10 s

acceleration = 0.5 m/s²

speed at which it crosses the finish line = ?

v = 36 x 0.278 = 10 m/s

using equation of motion

v = u + a t

v = 10 + 0.5 x 10

v = 15 m/s

v = 15 x 3.6 = 54 km/hr

speed at which the biker crosses the finish line is equal to 54 km/h

4 0

3 years ago