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4 years ago

At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range?

Physics

1 answer:

miss Akunina [59]4 years ago

8 0

The water should be aimed 45 deg above the horizontal.

That's half-way up from the horizon to the point directly over your head.

Same goes for tossing a stone or a baseball for maximum distance.

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When a +0.00235 C charge

irakobra [83]

Answer:177.87

Explanation:

4 0

3 years ago

A cart of mass M is attached to an ideal spring that can stretch and compress equally well. The cart and spring rest on a smooth

Alexxx [7]

We have to remember a point , which is ' the cart or spring rest on a smooth horizontal track ' , i.e., any frictional force doesn't take place.

Explanation:

a) According to the question the cart is pulled to position A and released, i.e., the velocity of the cart at A initially (say time,t=0) is 0 m/s ,then moves toward position E, where it reverses direction and returns again to position A , in the 2nd phase cart moves along A to E , the cart's velocity increase and again goes to zero at point E and again change the direction, hence

( File has been attached)

b) Let's , the distance between two consecutive points is x meter and the spring constant is k N.m

c) ( File has been attached)

d) Movinf Right

e) Moving left

4 0

3 years ago

The escape speed from a very small asteroid is only 24 m/s. If you throw a rock away from the asteroid at a speed of 30 m/s, wha

svp [43]

Given Information:

Initial speed of rock = vi = 30 m/s

escape speed of the asteroid = ve = 24 m/s

Required Information:

final speed of rock = vf = ?

Answer:

vf = 18 m/s

Explanation:

As we know from the conservation of energy

KEf + Uf = KEi + Ui

Where KE is the kinetic energy and U is the potential energy

0 + 0 = ½mve² - GMm/R

When escape speed is used, KEf is zero due to vf being zero. Uf is zero because the object is very far away from mass M, therefore, the equation becomes

GMm/R = ½mve²

m cancels out

GM/R = ½ve²

GM/R = ½(24)²

GM/R = 288

KEf + Uf = KEi + Ui

½mvi² + 0 = ½vf² - GMm/R

m cancels out

½vi² = ½vf² - GM/R

Substitute the values

½(30)² = ½vf² - (288)

½vf² = 450 - 288

vf² = 2(162)

vf = √324

vf = 18 m/s

Therefore, the final speed of the rock is 18 m/s

8 0

3 years ago

Read 2 more answers

Please! Help me (with explanation)

torisob [31]

B. The angle will cause the suns rays to be reflected downward to the other scout

4 0

3 years ago

a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe

Nikitich [7]

Answer:

(A) The spring constant is 98 N/m.

(B) The period of oscillation is 0.635 s.

Explanation:

Given:

Mass of the body is, $m=1\ kg$

Extension length of the spring is, $x=10\ cm=0.1\ m$

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

$F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N$

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

$F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m$

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

$T=2\pi\sqrt{\frac{m}{k}}$

Plug in the given values and solve for period 'T'. This gives,

$T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s$

Therefore, the period of oscillation is 0.635 s.

5 0

4 years ago