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3 years ago

Will anyone please help me solve this physics Problem?

Physics

1 answer:

ArbitrLikvidat [17]3 years ago

5 0

Answer:

I. Clockwise moment = 150 Nm

II. Anticlockwise moment = 400 Nm

III. 200 N

III. The anticlockwise moment.

Explanation:

From the question given above, we can obtain the answers to the questions as follow:

I. Determination of the clockwise moment.

Distance (d) = 1.5 m

Force (F) = 100 N

Clockwise Moment =?

Moment = Force × distance.

Clockwise moment = 100 × 1.5

Clockwise moment = 150 Nm

II. Determination of the anticlockwise moment.

Distance (d) = 2 m

Force (F) = 200 N

Anticlockwise Moment =?

Moment = Force × distance.

Anticlockwise moment = 200 × 2

Anticlockwise moment = 400 Nm

III. Determination of who will turn the sea saw.

Clockwise moment = 150 Nm

Anticlockwise moment = 400 Nm

Since the anticlockwise moment is greater than the clockwise moment, it therefore means that the anticlockwise moment will turn the sea saw.

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A ball of mass m on a string of length L is attached to a pivot. The ball is released from rest while the string is parallel to

mash [69]

Answer:

L/2

Explanation:

Neglect any air or other resistant, for the ball can wrap its string around the bar, it must rotate a full circle around the bar. This means the ball should be able to swing to the top position where it's directly above the bar. By the law of energy conservation, this happens when the ball is at the same level as where it's previously released vertically. It means the swinging radius around the bar must be at least half of the string length.

So the distance d between the bar and the pivot should be at least L/2

8 0

3 years ago

luda_lava [24]

Answer:

500000N/m²

5250N

Explanation:

Given parameters:

Depth(H) = 50m

Density of water = 1000kg/m³

Acceleration of free fall = 10m/s

Unknown:

Pressure the water exerts on the diver = ?

Solution:

Pressure is the force per unit area on a body. In fluids, pressure is the product of density, gravity and height

Pressure in fluids = Density x acceleration due to gravity x height

Input the variables and solve;

Pressure in fluids = 1000 x 10 x 50 = 500000N/m²

width of window = 150mm

height of window = 70mm

Force water exerts on the window = ?

To solve this problem;

Pressure = $\frac{Force}{Area}$

Area of the window = width x height = 150 x 10⁻³ x 70 x 10⁻³

= 1.05 x 10 ⁻²m²

Force = pressure x area

Input the variables;

= 500000N/m² x 1.05 x 10 ⁻²m²

= 5250N

4 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago

Given two vectors A=4i^+3j^ and vector B=5i^-2j^.find the magnitude of each vector

Flura [38]

Answer:

Magnitude of A=5

Magnitude of B=5.39

Explanation:

The magnitude of Vectors in Rectangular Form

Given a vector v in its rectangular form:

$\mathbf{v}=x\hat i+y\hat j$

The magnitude of v is:

$\mid\mid\mathbf{v}\mid \mid=\sqrt{x^2+y^2}$

We are given the vectors

$\mathbf{A}=4\hat i+3\hat j$

$\mathbf{B}=5\hat i-2\hat j$

Their magnitudes are:

$\mid\mid\mathbf{A}\mid \mid=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}$

$\mid\mid\mathbf{A}\mid \mid=5$

$\mid\mid\mathbf{B}\mid \mid=\sqrt{5^2+(-2)^2}=\sqrt{25+4}=\sqrt{29}$

$\mid\mid\mathbf{B}\mid \mid=\sqrt{29}=5.39$

4 0

3 years ago

An object has a starting velocity of 20 m/s. If it accelerates at a rate of 2 m/s' for 3 seconds,

Ghella [55]

Answer:

26 m/s

69 m

Explanation:

Given:

v₀ = 20 m/s

a = 2 m/s²

t = 3 s

Find: v and Δx

v = at + v₀

v = (2 m/s²) (3 s) + 20 m/s

v = 26 m/s

Δx = v₀ t + ½ at²

Δx = (20 m/s) (3 s) + ½ (2 m/s²) (3 s)²

Δx = 69 m

3 0

3 years ago

Will anyone please help me solve this physics Problem?​

Will anyone please help me solve this physics Problem?