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Dima020 [189]
2 years ago
14

"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of

water is 1000 kg/m3."
Physics
2 answers:
Alja [10]2 years ago
6 0

Answer:

buoyant force on the block due to the water= 10 N

Explanation:

We know that

buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.

Given:

A block of metal weighs 40 N in air and 30 N in water.

F_B =  40-30= 10 N

therefore,  buoyant force on the block due to the water= 10 N

Yuliya22 [10]2 years ago
3 0

Answer:

10 N

Explanation:

We are given that

Weight of block of metal in air=40 N

Actual weight of object=Weight of object in air

Actual weight of block of metal=40 N

Apparent weight of block of metal=Weight of block in water

Weight of block in water=30 N

Apparent weight of block=30 N

Density of water=1000 kg /cubic meter

We have to find the buoyant force on the block due to water.

Buoyant force=Actual weight-apparent weight

Buoyant force on the block due to water=40-30=10 N

Hence, the buoyant force on the block due to the  water=10 N

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evaluate the numerical value of the vertical velocity of the car at time t=0.25 s using the expression from part d, where y0=0.7
likoan [24]

Given :

Displacement , y = 0.75 m .

Angular acceleration , \alpha=0.95\ s^{-2} .

Initial angular velocity , \omega_o=6.3\ s^{-1} .

To Find :

The value of vertical velocity after time t = 0.25 s .

Solution :

By equation of circular motion is given by :

\omega=\omega_o+\alpha t

Putting all given values we get :

\omega=6.3+0.95\times 0.25\\\\\omega= $$6.5375\ s^{-1}

Now , vertical velocity is given by :

v=y\omega\\\\v=0.75\times 6.5375\ m/s\\\\v=4.90\ m/s

Therefore , the numerical value of the vertical velocity of the car at time t=0.25 s is 4.90 m/s .

Hence , this is the required solution .

8 0
2 years ago
A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it i
il63 [147K]

Answer:

a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to

This can be calculated by Gauss' Law.

A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.

b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>

8 0
2 years ago
Which is a better conductor, a flagpole or a flag? Why?
Novosadov [1.4K]

Answer:

flagpole

Explanation:

if it is about electricity then its flagpole

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2 years ago
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The grooved pulley of mass m is acted on by a constant force F through a cable which is wrapped securely around the exterior of
Sidana [21]

Answer:

Answer; v= 1.2654m/s

T= 110.76N

Explanation:

Apply Momentum Principle

Fdtro - Mgridt = Iow +Mvr

Fdtro - Mgridt = mK2 v/r1 + Mvr1

85 x 3x 0.345 -11 x 9.81 x 0.23 x 3 =30 x 0.25 x 0.25 x v/0.23 + 11 x v x 0.23 =

v = 1.2654m/s

To find the timed average value

Tdt -Mgdt =MV

T x 3 - 11 x 9.81 x 3 = 11 x 0.778

T= 110.76N

3 0
2 years ago
When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscil
marishachu [46]

Answer:

a) 40,000 N/m

b) f = 6.37 Hz

c) v = 4,8 m/s

Explanation:

part a)

First in order to estimate the spring constant k, we need to know the expression or formula to use in this case:

k = ΔF / Δx

Where:

ΔF: force that the men puts in the car, in this case, the weight.

Δx: the sinking of the car, which is 2 cm or 0.02 m.

With this data, and knowing that there are four mens, replace the data in the above formula:

W = 80 * 10 = 800 N

This is the weight for 1 man, so the 4 men together would be:

W = 800 * 4 = 3200 N

So, replacing this data in the formula:

k = 3200 / 0.02 = 160,000 N/m

This means that one spring will be:

k' = 160,000 / 4 = 40,000 N/m

b) An axle and two wheels has a mass of 50 kg, so we can assume they have a parallel connection to the car. If this is true, then:

k^n = 2k

To get the frequency, we need to know the angular speed of the car with the following expression:

wo = √k^n / M

M: mass of the wheel and axle, which is 50 kg

k = 40,000 N/m

Replacing the data:

wo = √2 * 40,000 / 50 = 40 rad/s

And the frequency:

f = wo/2π

f = 40 / 2π = 6.37 Hz

c) finally for the speed, we have the time and the distance, so:

V = x * t

The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:

x: distance of 80 cm or 0.8 m

V = 0.8 * 6 =

V = 4.8 m/s

5 0
2 years ago
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