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2 years ago

Which is more dense lead or Gold ?

Physics

2 answers:

WARRIOR [948]2 years ago

8 0

Answer:

Gold is more dense.

Gold sinks faster than lead. That's why gold is found in the bottom of a gold pan or river.

never [62]2 years ago

4 0

Answer:

Gold is much heavier than lead. It is very dense. ... Therefore gold weighs 19.3 times as much or (19.3 x 8.3 lb) about 160 pounds per gallon. Although gold has a density 19.3 times greater than water and is one of the most dense substances on Earth, there are substances with far more amazing densities.

Explanation:

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A crowbar with the length of 200 cm is used to lift the load of 600 Newton . If the distance between fulcrum and load is 0.75 .

Leviafan [203]

Answer:

a. Effort = 960 Newton

b. Mechanical advantage (M.A) = 0.625

c. Velocity ratio (V.R) = 1.67

Explanation:

Given the following data;

Load = 600 N

Length of crowbar = 200 cm

Length of load arm = 0.75 m

Conversion:

100 cm = 1 m

X cm = 0.75 m

Cross-multiplying, we have;

X = 0.75 * 100 = 75 cm

First of all, we would find the effort arm;

Effort arm = length of crow bar - length of load arm

Effort arm = 200 - 75

Effort arm = 125 cm

Next, we would determine the mechanical advantage (M.A) of the crow bar;

Substituting the values into the formula, we have;

M.A = 0.625

To find the effort of the crow bar;

Making "effort" the subject of formula, we have;

Effort = 960 Newton

Lastly, we would determine the velocity ratio (V.R);

V.R = 1.67

8 0

1 year ago

Which title best reflects the main idea of the passage? The Role of Convection in the Distribution of Earth's Energy The Role of

Leto [7]

Answer:

The Role of Heat Transfer Methods in the Distribution of Earth's Energy

Explanation:

8 0

3 years ago

Read 2 more answers

A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes. If the club and ball are in contact fo

LuckyWell [14K]

Answer:

Average force, F = 562.5 N

Explanation:

Mass of the golf ball, m = 0.045 kg

Initially, it is at rest, u = 0

Final speed of the ball, v = 25 m/s

The club and the ball are in contact for, $t=2\ ms=2\times 10^{-3}\ s$

We need to find the average force acting on the ball. It can be calculated using the formula as :

$F\times t=m(v-u)$

$F=m\dfrac{v-u}{t}$

$F=0.045\times \dfrac{25}{2\times 10^{-3}}$

F = 562.5 N

So, the average force acting on the ball is 562.5 N. Hence, this is the required solution.

5 0

3 years ago

An air-plane has an effective wing surface area of 17.0 m² that is generating the lift force. In level flight the air speed over

Lady bird [3.3K]

Answer:

Explanation:

Given that,

Surface area A= 17m²

The speed at the top v" = 66m/s

Speed beneath is v' =40 m/s

The density of air p =1.29kg/m³

Weight of plane?

Assuming that,

the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.

Using Bernoulli equation

P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''

Where

P' is pressure at the bottom in N/m²

P" is pressure at the top in N/m²

v' is velocity at the bottom in m/s

v" is velocity at the top in m/s

Then, Bernoulli equation becomes

P'+ ½pv'² = P'' + ½pv''²

Rearranging

P' — P'' = ½pv"² —½pv'²

P'—P" = ½p ( v"² —v'²)

P'—P" = ½ × 1.29 × (66²-40²)

P'—P" = 1777.62 N/m²

Lift force can be found from

Pressure = force/Area

Force = ∆P ×A

Force = (P' —P")×A

Since we already have (P'—P")

Then, F=W = (P' —P")×A

W = 1777.62 × 17

W = 30,219.54 N

The weight of the plane is 30.22 KN

5 0

3 years ago

Read 2 more answers

A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down

Len [333]

Answer: 0m/s²

Explanation:

Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane

According newton's law of motion

Summation of forces along the plane = mass × acceleration

Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

Ff = nR where

n is coefficient of friction = tan(theta)

R is normal reaction = Wcos(theta)

Fm = Wsin(theta)

Substituting in the formula of newton's first law we have;

Fm-Ff = ma

Wsin(theta) - nR = ma

Wsin(theta) - n(Wcos(theta)) = ma... 1

Given

W = 562N, theta = 30°, n = tan30°, m = 56.2kg

Substituting in eqn 1,

562sin30° - tan30°(562cos30°) = 56.2a

281 - 281 = 56.2a

0 = 56.2a

a = 0m/s²

This shows that the trunk is not accelerating

4 0

3 years ago