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artcher [175]
2 years ago
14

A dart player throws a dart horizontally at a velocity of 12.7 m/s. The

Physics
1 answer:
Igoryamba2 years ago
3 0

Answer:

Answer

Explanation:

It is given that,

The horizontal speed of a dart player =

12.4 m/s

The dart hits the board 0.32 m below the

height from which it was thrown.

We need to find how long it take the dart

to hit the board.

Let the time taken by the dart to hit the

board. Using second equation of motion

to find it.

1

y = ut +

2

=

at2

Put u = 0 and a=-g

-

-1

y =

1942

1

-0.32

=

x 9.8t2

2

2 x 0.32-0.32

=

t=

9.8

t= 0.255 s

Let x be the horizontal distance travelled.

It is calculated as follows:

X =

12.4 x 0.255

=

= 3.16 m

So, the dart will hit the board at a

distance of 3.16 m.

Muscardinus • Quality

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The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a
vampirchik [111]

Answer:

F = 1.5 \times 10^{-16} N

this force is 1.68 \times 10^{13} times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

KE = 1 keV

KE = 1 \times 10^3 (1.6 \times 10^{-19}) J

KE = 1.6 \times 10^{-16} J

now the speed of electron is given as

KE = \frac{1}{2}mv^2

now we have

v = \sqrt{\frac{2 KE}{m}}

v = 1.87 \times 10^7 m/s

now the maximum force due to magnetic field is given as

F = qvB

F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})

F = 1.5 \times 10^{-16} N

Now if this force is compared by the gravitational force on the electron then it is

\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}

\frac{F}{F_g} = 1.68 \times 10^{13}

so this force is 1.68 \times 10^{13} times more than the gravitational force

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If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gr
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A thin 1.5 mm coating of glycerine has been placed between two microscope slides of width 0.8 cm and length 3.9 cm . Find the fo
Radda [10]

The  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

<h3>Force required to pull one end at a constant speed</h3>

The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;

F = ma

where;

  • m is mass
  • a is acceleration

At a constant speed, the acceleration of the object will be zero.

F = m x 0

F = 0

Thus, the  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

Learn more about constant speed here: brainly.com/question/2681210

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1 year ago
Question 3(Multiple Choice Worth 3 points)
Burka [1]

The unit of measurement used in the metric system is  Meters.

The metric system makes use of measurements that we often see every day. these are used to measure length, weight, and capability. you will often see these as grams, kilograms, millimetres, centimeters, meters, milliliters, and liters. The metric machine is used all around the globe as it is easy to understand.

The metric system is a system of measurement that succeeded the decimalized system primarily based on the meter that had been added in France in the 1790s.

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8 0
1 year ago
Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).
julia-pushkina [17]

Answer:

A=6.28\ \mu m^2

Part 1

V=1.57 \ \mu m^3

Part 2

A=6.28\ \mu m^2

Explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

V=\pi r^2l

V=\pi \times 0.5^2\times 2 \ \mu m^3

V=1.57 \ \mu m^3

Surface area,A

A=π d l

A=\pi \times 1 \times 2\ \mu m^2

A=6.28\ \mu m^2

Part 1

V=1.57 \ \mu m^3

Part 2

A=6.28\ \mu m^2

4 0
3 years ago
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