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3 years ago

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A car is traveling at a speed of 54 km/h. Breaks are applied so as to produce a uniform acceleration of -0.5 m/s2. Find how far

the car will go before it is brought to rest

1 answer:

Lena [83]3 years ago

8 0

Explanation:

u=54 km/h

54*5/18=15 m/s

v=0m/s

t=?

acceleration=-0.5m/s^2

we know that a=v-u/t

so,

t=v-u/a

t=15-0/0.5

=15/0.5

=30

therefore, the time is 30 second

Hope this answer helps you..

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A laboratory electromagnet produces a magnetic field of magnitude 1.50T . A proton moves through this field with a speed of 6.00

hram777 [196]

The magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

The magnitude of the force on a charged particle moving in a magnetic field is given by the formula,

F= qvB

Here q is the charge on proton = 1.6 x 10^-19 C.

v is the velocity with which the particle is moving = 6.00 x 10^6 m/s

And B is the value of the magnetic field = 1.5 T

Putting the given values in the above equation,

F = 1.6 x 10^-19 x 6 x 10^6 x 1.5 = 1.44 x 10^-12 N.

Hence, the magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

To know more about "magnetic force", refer to the link given below:

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1 year ago

Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral

cluponka [151]

Answer:

F = μi²l/πa

Explanation:

The magnetic force F on a length of wire, l carrying a current i in a magnetic field B is given by

F = Bilsinθ

The magnetic field due to one wire of length, l carrying a current, i at a distance a from it is given by B = μi/2πa

So, the force on the first wire due to the second wire is F₁ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

the force on the first wire due to the third wire is F₂ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

Since the magnetic field due to the one wire is perpendicular to the length of the other wire its field acts upon, θ = 90

So, F₁ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa and

F₂ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa

Since the angle between F₁ and F₂ is 60° (since it is an equilateral triangle)

The resultant force F is thus

F = √(F₁² + F₂² + 2F₁F₂cos60°)

F = √(F₁² + F₂² + 2F₁F₂ × 0.5)

F = √(F₁² + F₂² + 2F₁F₂) (since F₁ = F₂)

F = √(2F₁² + 2F₁²) = √(4F₁²)

F = 2F₁

F = 2μi²l/2πa

F = μi²l/πa

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3 years ago

Read 2 more answers

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

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3 years ago

In levelling, the following staff readings were observed involving an inverted staff, A = 2.915 and B = -2.028. What is the rise

salantis [7]

Answer:

The rise from A to B is 0.887

Solution:

As per the question:

The following reading of an inverted staff is given as:

A = 2.915

B = -2.028

Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.

Thus

The rise from A to B is given as:

A - B = 2.915 - 2.028 = 0.887

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3 years ago

In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s

serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

$a = \dfrac{1.2 \times 10^5}{86}$

$a =1395.35\ m/s^2$

Using equation of motion

v² = u² + 2 a s

$s =\dfrac{v^2}{2a}$

$s =\dfrac{52^2}{2\times 1395.35}$

s = 0.9689 m

The minimum depth of snow that would have stooped him is s = 0.9689 m

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3 years ago