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2 years ago

hey pls help with this HDJSJDH im rlly bad at this kinda stuff :,) i'll mark the correct answer brainliest

Physics

2 answers:

Mama L [17]2 years ago

8 0

Answer:

A. Cue, White Ball, Blue Ball

Explanation:

Visualize in which order things are being hit

timama [110]2 years ago

7 0

I am pretty sure it’s A
The cue exerts force onto the white ball which pushes the blue ball into the direction of the hole.

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An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m

Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

$\omega = \sqrt{\frac{k}{m}}$

here we know

m = 3.5 kg

k = 270 N/m

now we have

$\omega = \sqrt{\frac{270}{3.5}}$

$\omega = 8.78 rad/s$

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

$v = \omega \sqrt{A^2 - x^2}$

$0.55 = 8.78 \sqrt{A^2 - 0.020^2}$

$A = 0.066 m$

Part b)

Maximum speed of SHM at its mean position is given as

$v_{max} = A\omega$

$v_{max} = 0.066(8.78) = 0.58 m/s$

4 0

3 years ago

A vertical, solid steel post 25 cm in diameter and 2.50m long is required to support a load of 8000kg. You can ignore the weight

Gwar [14]

(a) The stress in the post is 1,568,000 N/m²

(b) The strain in the post is 7.61 x 10⁻⁶

<h3>Area of the steel post</h3>

A = πd²/4

where;

d is the diameter

A = π(0.25²)/4 = 0.05 m²

<h3>Stress on the steel post</h3>

σ = F/A

σ = mg/A

where;

m is mass supported by the steel
g is acceleration due to gravity
A is the area of the steel post

σ = (8000 x 9.8)/(0.05)

σ = 1,568,000 N/m²

<h3>Strain of the post</h3>

E = stress / strain

where;

E is Young's modulus of steel = 206 Gpa

strain = stress/E

strain = (1,568,000) / (206 x 10⁹)

strain = 7.61 x 10⁻⁶

<h3>Change in length of the steel post</h3>

strain = ΔL/L

where;

ΔL is change in length
L is original length

ΔL = 7.61 x 10⁻⁶ x 2.5

ΔL = 1.9 x 10⁻⁵ m

Learn more about Young's modulus of steel here: brainly.com/question/14772333

#SPJ1

7 0

1 year ago

Pls help. This is about heat. I got the answer, just need the working

Mars2501 [29]

Answer:

See below

Explanation:

You have to heat the calorimeter to 100 C from 20 C

this will take .20 kg * 390 j /kg-C * 80 C = <u>6240 j</u>

You have to heat the mass of water to boiling point (100 C ) from 20C

this will take

.50 kg * 4182 j/kg-C * 80 = <u>167,280 j </u>

AND you have to add enough heat to boil off .03 kg of water:

.03 kg * (2260000 j/kg-C ) =<u> 67,800 j</u>

<u />

Power = joules / sec = (6240 + 167280 + 67800) / 274.8 =<u> 878 watts </u>

<u />

<u>Your answer may differ just a bit for slightly different or rounded values of specific heat or heat of fusion for water .....</u>

7 0

2 years ago

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

q₁ = + 1.25 nC

3 0

3 years ago

About light and energy

Sophie [7]

E = MC^2. Albert Einstein's proven formula. When mass travels at the square of speed of light, the mass gets converted into energy

7 0

3 years ago