Calculate the acceleration of the body of mass 3kg on which a force of 42N has been applied for 5s

if it takes 3.5 hours for the hogwarts express moving at a speed of 120 mi/hr to make it from platform 9 and 3/4 to hogwarts how

Mila [183]

Use the formula distance = rate * time. substitute in the values you know, which gives you distance = 120mph * 3.5hours. multiply 120 and 3.5 to get distance = 420 miles.

4 0

2 years ago

vector A makes equal angles with x,y and z axis. value of its components (in terms of magnitude of vector A will be?

kiruha [24]

X^2+y^2+z^2=A^2
But here XY and Z are all equal so
3X^2=A^2
X=A/(sqrt(3))
Each component is the value of a divided by the square root of three. This way if you square then and add them up it equals a squared

5 0

2 years ago

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A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for

Lorico [155]

The force tending to lift the load (vertical force) is equal to 22.5N.

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

$VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N$

Hence, we can see that the force tending to lift the load off the ground (vertical force) is equal to 22.5N.

Have a nice day!

8 0

2 years ago

A densly wound cylindrical coil has 210 turns per meter, a 5 cm radius, and carries 38 mA. What is the magnitude of the uniform

kaheart [24]

Answer:

(a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

Explanation:

Given that,

Number of turns = 210

Radius = 5 cm

Current = 38 mA

Current in the wire = 500 mA

We need to calculate the magnetic field inside a coil

Using formula of magnetic field

$B=\mu_{0} ni$

Put the value into the formula

$B_{c}=4\pi\times10^{-7}\times210\times38\times10^{-3}$

$B_{c}=0.0000100\ T$

$B_{c}=10.02\times10^{-6}\ T$

Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil

Distance $d=\dfrac{5}{2}$

$d=2.5\ cm$

We need to calculate the magnetic field from the straight wire

Using formula of magnetic field

$B_{w}=\dfrac{\mu_{0}I}{2\pi d}$

Put the value into the formula

$B_{w}=\dfrac{4\pi\times10^{-7}\times500\times10^{-3}}{2\times\pi\times2.5\times10^{-2}}$

$B_{w}=4.0\times10^{-6}\ T$

This field is perpendicular to the wire.

The magnitude of magnetic field is

$B=\sqrt{B_{c}^2+B_{w}^2}$

$B=\sqrt{(10.02\times10^{-6})^2+(4.0\times10^{-6})^2}$

$B=0.00001078\ T$

$B=10.78\times10^{-6}\ T$

$B=10.78\ \mu\ T$

Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

6 0

2 years ago

A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a

JulijaS [17]

All the weight of the wooden board is bear by the support located at the centre of the rod, and the other support which is located at the end, will have no reaction force, or 0 reaction force.

Therefore the reaction at the centre support is equal to the weight of the board, while the support at the end has 0 reaction force.

8 0

2 years ago

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