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2 years ago

The maximum value of magnetic in an electric field 3.2 *10^4

Physics

1 answer:

Svetradugi [14.3K]2 years ago

6 0

Answer:

the answer is 12 because if your magnetic value and Electric field is 3.2 the answer will be 12

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A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an x axis. He can legally

Over [174]

Answer:

minimum angle is 128.69°

Explanation:

given data

player velocity with respect ground v1 = 3.5 m/s

ball velocity with respect himself v2 = 5.6 m/s

to find out

smallest angle

solution

we know ball velocity with respect field will be

ball velocity = v1 +v2

ball velocity = 3.5 + 5.6 = 9.1m/s

we consider angle that player hit ball is θ

then by as per figure triangle

cosθ = $\frac{v1}{v2}$

cosθ = $\frac{3.5}{5.6}$

θ = 51.31

so minimum angle is 180 - 51.31 = 128.69°

6 0

3 years ago

A 900 kg car moves around a 500 m radius curve at 25.0 m/s. What is the centripetal force on

Gnesinka [82]

Answer:

9375 N

Explanation:

From the question,

Centripetal force (F) = mv²/r.................. Equation 1

Where m = mass of the car, v = velocity of the car, r = radius of the curve.

Given: m = 900 kg, r = 600 m, v = 25 m/s

Substitute these values into equation 1

F = (900×25²)/600

F = 9375 N.

Hence the centripetal force on the car is 9375 N

3 0

2 years ago

A spring that is compressed 14.5 cm from its equilibrium position stores 2.99 J of potential energy. Determine the spring consta

strojnjashka [21]

Answer:

284.4233 N/m

Explanation:

k = Spring constant

x = Compression of spring = 14.5 cm

U = Potential energy = 2.99 J

The potential energy of a spring is given by

$U=\dfrac{1}{2}kx^2$

Rearranging to get the value of k

$\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 2.99}{0.145^2}\\\Rightarrow k=284.4233\ N/m$

The spring constant is 284.4233 N/m

7 0

3 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

7 0

3 years ago

PLS HELP !!

Ira Lisetskai [31]

A because it’s basic kinetic example

4 0

2 years ago

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The maximum value of magnetic in an electric field 3.2 *10^4​

The maximum value of magnetic in an electric field 3.2 *10^4