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Black_prince [1.1K]
2 years ago
6

A vertical spring launcher is attached to the top of a block and a ball is placed in the launcher, as shown in the figure. While

the block slides at constant speed to the right across a horizontal surface with negligible friction between the block and the surface, the ball is launched upward. When the ball reaches its maximum height, what will be the position of the ball relative to the spring launcher?.
Physics
1 answer:
kvasek [131]2 years ago
7 0

For a vertical spring launcher is attached to the top of a block and a ball is placed in the launcher, the position of the ball will be  behind the box

<h3>What will be the position of the ball relative to the spring launcher?</h3>

Generally, the equation for the  conservation of momentum principle  is mathematically given as

(M+m) V1 = M*V2

Therefore, with the ball moving forward we have that; the ball at top it wii be behind the box,

Read more about Motion

brainly.com/question/605631

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Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
3 years ago
Q¹=0,07Mc Q²=2C r=1,08 cm F=.......?
Nostrana [21]
The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2)  where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.

Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span> 
5 0
3 years ago
QC In ideal flow, a liquid of density 850 kg / m³ moves from a horizontal tube of radius 1.00cm into a second horizontal tube of
kvv77 [185]

The  volume flow rates for ∆P is 6.81m³/s .

<h3>What is pressure?</h3>

The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.

According to bernaulli's theorem theorem

P+1/2pV²+pgy = constant

where p fluid density

g is acceleration due to gravity, pressure at elevation,v is Velocity at elevation ,y is height of elevation.

As there are two tubes then the height of tube 1 is equal to height of tube two .

P1-P2=1/2p(Vd²-Vl²)

The flow rate of liquid is  A1V1=A2V2 .

rest is attached in image

to learn more about Pressure click here brainly.com/question/12971272

#SPJ4

4 0
2 years ago
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how ma
Alenkinab [10]

Answer:

The wavelength stays the same.

Explanation:

When the amplitude is increased, the wavelength stays the same.

Here the wavelength doesn't depend upon the amplitude.

4 0
3 years ago
In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?
IceJOKER [234]

Answer: a=1016.66 m/s^{2}

Explanation:

Acceleration a is expressed in the following formula:

a=\frac{V_{f}-V_{o}}{t}

Where:

V_{f}=610 m/s is the final velocity of the projectile

V_{o}=0 m/s  is the initial velocity of the projectile

t=0.6 s is the time

Solving:

a=\frac{610 m/s-0 m/s}{0.6 s}

a=1016.66 m/s^{2} This is the acceleration of the projectile

6 0
3 years ago
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