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3 years ago

How to get more water pressure in shower?

1 answer:

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<span> I picked up cost $1.50). If you look down inside of the showerhead you'll see a washer which constricts the water flow. If you take a drill and enlarge the hole, thereby increasing flow, the water </span>pressure<span> will increase</span>

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Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass

Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

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3 years ago

starting from a stop a traffic signal, a car speeds up to 20 m/s in 5 seconds. calculate the acceleration of the car.

Ira Lisetskai [31]

Answer:

5.867 m/s^2

Explanation:

Initial Speed > 0 m/s

Final Speed > 20 m/s

Time > 5 sec.

8 0

3 years ago

- Name two elements that have the SAME (rounded) atomic mass:

katovenus [111]

Answer:

Magnesium atomic no. = 24,25,26. These are the two elements which have same atomic no

Explanation:

3 0

2 years ago

The question is in the picture

Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

$v_0-gt = 0$ .

where $v_0$ is the initial velocity.

Solving for $t$ we get

$t = \dfrac{v_0}{g}$

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time $t_0$ since it had reached the highest point, the ball has traveled downwards and the velocity $v_f$ it has gained is

$v_f = gt_0$ ,

and we are told that this is twice the initial velocity $v_0$ ; therefore,

$v_f = 2v_0 = gt_0$

which gives

$t_0 = \dfrac{2v_0}{g}.$

Thus, the total time taken to reach velocity $2v_0$ is

$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

$36= \dfrac{3v_0}{g},$

and solving for $v_0$ we get:

$v_0 = \dfrac{36g}{3}$

$v_0 = \dfrac{36s(10m/s^2)}{3}$

$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

7 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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2 years ago