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3 years ago

PLEASE ANYONE CAN HELP ME !E.x/A block of metal has a volume of 0.09 m3

Physics

1 answer:

LuckyWell [14K]3 years ago

3 0

Answer:

B = 1058.4 N

Explanation:

Given that,

The volume of a metal block, V = 0.09 m³

The density of fluid, d = 1200 kg/m³

We need to find the buoyant force when it's Completely immersed in brine. The formula for the buoyant force is given by :

$B=\rho gV$

g is acceleration due to gravity

$B=1200\times 9.8\times 0.09\\\\B=1058.4\ N$

So, the required buoyant force is 1058.4 N.

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Need help!!! Physics due tomorrow help, help ,help

Sergio [31]

1. Due to the first law of Newton: there is a force that accelerate the body,

then the net force for 2 forces in opposite directions can calculated as

$F_{net}=F_{1}-F_{2}=26-8=14N$

2. For a force make an angle with the horizontal plane, we need to analysis the force into 2 perpendicular forces:

- In horizontal plane:

$F_{x}=Fsin ($ θ $)= 67sin(73)=64N$

then, as Q(1),

$F_{net}=F_{1}-F_{2}=64-33=31 N$

3. Because of the non change in the velocity, there is no acceleration

Then, there is no force is acted from the body

Hence all forces is according to the tension in the cable

$F_{Tension}=420 N$

4. A) As we solve the Q(2), we will analysis the un horizontal or vertical forces into 2 forces

$F_{x}=Fsin ($ θ $)= 600sin(45)=424.264N$

$F_{y} = F sin ($ θ $)= 600cos45)=424.264N$

Hence we have 2 opposite horizontal forces. also, 2 opposite vertical forces.

now, we calculate the net forces in each direction

$F_{net in x}=F_{1}-F_{2}=424.264-250=174.264 N$ in east

$F_{net in y}=F_{1}-F_{2}=-500-424.264=75.736 N$ in south

The total net force for perpendicular forces is calculated from the relation:

$F_{total}=\sqrt{F_{x} ^{2}+F_{y}^{2}}=\sqrt{174.264^{2}+75.736^{2}}=190N$

B) To find the direction, we need to calculate the angle which the total force is made with the east direction of the x-axis (horizontal)

θ = $tan^{-1}(\frac{F_{y}}{F_{x} })= tan^{-1}(\frac{-75.736}{174.264})=-23.48^{o}$

That meas that the total force is made with the east direction of the x-axis angle equal to 23.48 for the south (the 4th quarter)

5. According to the third law of Newton, to make the balance the net force equal to zero (for each action there is a reaction with the same amplitude and in opposite direction)

Then, $F_{right}=F_{left}=55N$

7 0

4 years ago

A 2500‐kg vehicle traveling at 25 m/s can be stopped by gently applying the breaks for 20 seconds. What is the average force sup

katen-ka-za [31]

Momentum = (mass) x (speed)

Change in momentum = (force) x (time)

The initial momentum is (mass) x (speed) = 2500x 25 = 62,500 kg-m/s.

Since you want to stop the vehicle, that number is also the required change
in momentum ... you want the vehicle to wind up with zero momentum.

62,500 = (force) x (time) = 20 x force

Divide each side by 20 :

force = 62,500 / 20 = 3,125 newtons 

3 0

3 years ago

What compound is formed from a tight network of oppositely charged ions

Vsevolod [243]

Answer:

salt

Explanation:

7 0

3 years ago

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large

svet-max [94.6K]

Given:

• Mass, m = 0.200 mg

• Speed, v = 3.00 x 10³ m/s

• Time, t = 6.00 x 10^⁻⁸ s.

Let's calculate the force exerted.

Using the inpulse-momentum theroerm, we have:

impulse = change in momemntum

Where:

Impulse = force x time

change in momentum = mass x velocity.

Thus, we have:

$F\times6.00\times10^{-8}=(0.200\times10^{-6})\times(3.00\times10^3)$

Let's solve for the force F:

$\begin{gathered} F=\frac{(0.200\times10^{-6})(3.00\times10^3)}{6.00\times10^{-8}} \\ \\ F=10000N \end{gathered}$

Therefore, the force exterted is 10000 N.

ANSWER:'

10000 N

4 0

1 year ago

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.50 N is applied. A 0.600-kg particle res

gulaghasi [49]

Answer:

Spring constant, k = 283.33 N/m

Explanation:

Given that,

Force acting on the spring, F = 8.5 N

Stretching in the spring, x = 3 cm = 0.03 m

Let k is the spring constant of the spring. It can be calculated using Hooke's law as :

$F=-kx$

$k=\dfrac{F}{x}$

$k=\dfrac{8.5\ N}{0.03\ m}$

k = 283.33 N/m

So, the spring constant of the spring is 283.33 N/m. Hence, this is the required solution.

5 0

4 years ago