Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz

Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is

dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

$W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N$

The mass of the water accumulated in the bucket after 3.20s is:

$m_w= (0.20 \ L/s) ( 3.20)s$

$m _w=0.64 \ kg$

To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

$F = v ( \dfrac {dm}{dt} )$

$F = (8.4 \ m/s)( 0.200 \ L/s)$

F = 1.68 N

Finally, the reading scale is:

$F_{scale$ = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

6 0

3 years ago

Read 2 more answers

A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he

zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water, u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

8 0

3 years ago

After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average

lara [203]

Answer:

compression of spring is x = 0.12 m

Assumed k = 160,000 N/m ........ Truck's suspension system

Explanation:

Given:

- The mass of average person m_p = 69 kg

- Total number of persons n_p = 27

- The mass of each goat m_g = 15 kg

- The total number of goats n_g = 3

- The mass of each chicken m_c = 3 kg

- The total number of goats n_c = 5

- The total mass of bananas m_b = 25 kg

Find:

How much are the springs compressed?

Solution:

- Using equilibrium equation on the taptap in vertical direction:

F_net = F_spring - F_weight = 0

- Compute the force due to all the weights on the taptap:

F_weight = (n_p*m_p + n_g*m_g + n_c*m_c + m_b)*9.81

F_weight = (69*27 + 3*15 + 5*3 + 25)*9.81

F_weight = 19109.88 N

- The restoring force of a spring is given by:

F_spring = k*x

Where, k is the spring stiffness and x is the displacement:

F_weight = F_spring

19109.88 = k*x

x = 19109.88 / k

We need to assume the spring stiffness we will take k = 160,0000 N/m (trucks suspension systems). The value of the stiffness must be high enough to sustain a load of 1.911 tonnes.

x = 19109.88 / 160,000

x = 0.1194 m ≈ 0.12 m = 12 cm

- A compression of 12 cm seems reasonable for a taptap to carry 1.911 tonnes of load. Hence, the assumption of spring stiffness was reasonable. Hence, the compression of spring is x = 0.12 m.

8 0

3 years ago

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti

Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0

3 years ago

some people agree that wearing seat belts reduces the risk of injurt to people in the car. however they say that the risk to oth

leva [86]

Answer:

It would be hard to test scientifically since it's subjective and can only be proven true if you conducted some experimentations and observations.

4 0

3 years ago

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz​

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz