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vova2212 [387]
4 years ago
11

a projectile is fired from the ground with an initial horizontal velocity of 8 meters per second. its initial vertical velocityi

s 6 meters per second. whats is the "total/resultant" velocity of the projectile at its peak?

Physics
1 answer:
Stels [109]4 years ago
6 0

a projectile is fired from the ground with an initial horizontal velocity of 8 meters per second. its initial vertical velocityis 6 meters per second. whats is the "total/resultant" velocity of the projectile at its peak?


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Answer:

Explanation:

a) A is accurate because the half of the Moon that is facing the sun is it by the sun, and the other half is dark.

3 0
3 years ago
A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I
Lorico [155]

Answer:

3.13cm/s²

Explanation:

Given

Initial velocity u = 3.5cm/s

Final velocity v = 8.2cm/s

Time t = 1.5secs

Required

Acceleration of the cart a

To get that, we will use the equation of motion

v = u+at

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8.2 = 3.5+1.5a

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a = 4.7/1.5

a = 3.13cm/s²

Hence the acceleration to the cart is 3.13cm/s²

3 0
3 years ago
A steel beam that is 6.50 m long weighs 336 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
FinnZ [79.3K]

Answer:

Explanation:

When beam is balanced and not rotating with Suki standing on it , let reaction force on the supports be R₁ and R₂. Then

R₁ +R₂ = 336 + 590

= 929

Now the moment beam begins to tip , reaction on distant support R₁ = 0

only R₂ will exists on the support near to Suki.

Taking torque about this support of weight of beam acting from the middle point and weight of suki of 590N ,who is x distance from the support towards the other end.

336 x 1.5 = 590 x

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4 0
3 years ago
Give an example of a system whose mass is not constant.
Sloan [31]
A spinning top is the answer
8 0
3 years ago
A 20 cm long spring is attached to a wall. The spring stretches to a length of 22 cm when you pull on it with a force of 100 n.
kenny6666 [7]

Answer:

5000 N/m

Explanation:

Hooke's law for the spring is

F = k \Delta x

where here we have

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Solving the equation for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{100 N}{0.02 m}=5000 N/m

8 0
4 years ago
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