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3 years ago

Iron ball sinks in water but not in mercury. Why?

Physics

2 answers:

Zigmanuir [339]3 years ago

8 0

Answer:

mercury is more dense

Explanation:

Anvisha [2.4K]3 years ago

7 0

Explanation:

Iron ball sinks in water but not in mercury because of the difference in the density iron used in iron ball. The iron ball has more density than water and less than the mercury.

Hence, Iron ball sinks in water but not in mercury.

-TheUnknownScientist

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How much force is needed to stop a 4000 kg truck moving at 8 m/s in 0 2 seconds?

maksim [4K]

Answer:

32000 N

Explanation:

Force Calculater

8 0

3 years ago

You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of

alina1380 [7]

Answer:

The force constant ,I = 2394N/m

Explanation:

Given:

Weight of crate,Wg = 1470N

Theta = 22.0°

Kinetic friction,Fk= Fs(max) = 550N

Total length of ramp =8.0m

If y =0 at the bottom of the ramp

y1 = d Sin theta

y1 = 8 × Sin 22°

y1 = 3.0m

y2 = 0

V1=1.8m/s

V2 = 0

The equation combining the gravitational and elastic potential energy, and the work energy theorem is given by:

K1 + Ugrav1 + Uel1 + W = K2 + Uel1 ..eq1

Where KE is given by: 1/2mv^2

Gravitational potential energy is given by: Ugrav = mgy ...eq2

Elastic potential energy = Uel = 1/2Kx^2 ..eq3

The restoring force constant of a spring compressed by a distance x is given by:

Fx = Kx ..eq4

Workdone by a constant force is given by W = Fscostheta ...eq5

Where s = displacement

Theta = the angle between the force and the displacement

Work,W = Wf = 550 ×8 × cos 180°

W = -4400J

From the weight of the crate.mass is:

Wg/g = m

1470/9.8 = 150kg

K1 = 1/2 ×150×1.8^2 = 243m/s

The crate comes to rest at K2=0

Ugrav1 = 150 × 9.8 × 3 = 4410J

Ugrav2 = 150 × 9.8 x 0 = 0J

Uel1= 0 Spring at equilibrium

Substituting the values of the energies and work

253 + 4410 + 0 - 4400 = 0 + 0 + Uel1

Uel1 = 253J

Substituting into eq3

253 = 1/2 Kx^2 = 1/2 Kx(x)

Kx = 506/x

Since crate remains at rest,we use Newton's 2nd law

Fx = fs + Wsin theta

Fx = 550 + 1470 sin 22

Fx = 1100.7N

Substituting into eq4

Kx = 1100.7

X = 506/1100.7 = 0.46m

Kx = 1100.7

K = 1100.7/0.47

K = 2394N/m

5 0

3 years ago

Can someone plz help me with this

Elena-2011 [213]

1st Law: Objects that are in motion tend to stay in motion. This motion can change with external forces.

If you were to stop pedaling on bike while in motion, you will notice that you will keep moving. This is because a moving body (you) has inertia. If there wasn't any friction between the tires and the ground, between the axles and wheel, any air resistance, or any other force that acts against you, then you could be coasting indefinitely! 

2nd Law: Force is equal to the mass times acceleration. 

When you pedal, you are applying a force onto the pedal. This force is then translated through tension to apply torque onto the wheel. Turning the wheel will make you accelerate in the lateral direction. 

3rd Law: For every action, there is an equal and opposite reaction. 

Without this, you could pedal and pedal, but you will be not go anywhere! It is essentially the friction between the tires and the ground that propels you forward. If the ground did not apply to the tire the same amount of force that the tire was applying to the ground, the tire would not "catch" and no friction would be applied. And if there was no third law, the weight of you and your bike would "sink" into the ground because the ground would not be applying a normal force back onto you.

hope this helps and if you have any questions just hmu and ask :)

3 0

3 years ago

A table is moved using 60 N of force.

dem82 [27]

Answer:

15m

Explanation:

W=f×s

$s = \frac{w}{f} \\ s = \frac{900j}{60n} \\ s = 15m$

5 0

3 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

3 years ago