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victus00 [196]
1 year ago
15

What is the net force in this image?

Physics
1 answer:
Lynna [10]1 year ago
8 0

Answer:

-15 as it inverse to gravity

the direction of big force

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Do all objects have inertia <br><br> a) true<br> b) false
Natasha2012 [34]

Answer:

The answer is A. Which is true

4 0
3 years ago
A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds af
disa [49]

Answer:

Explanation:

Given

Cannon is fired with a velocity of u=72.50\ m/s

Using Equation of motion

y=ut+\frac{1}{2}at^2

where

y=displacement

u=initial\ velocity

a=acceleration

t=time

after time t=3.3 s

y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2

y=239.25-53.36

y=185.89\ m

So after 3.3 s cannon ball is at a height of 185.89 m

6 0
3 years ago
A vector must always have both size and ..
DochEvi [55]
Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up
5 0
3 years ago
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
aivan3 [116]

Answer:

Part a)

F_n = 306 N

Part b)

v = 12.1 m/s

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

mg - F_n = ma_c

now we will have

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

now we have

F_n = 100(9.81) - \frac{100(9^2)}{12}

F_n = 981 - 675

F_n = 306 N

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

F_n + mg = ma_c

F_n = 0 at minimum speed

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15 \times 9.81}

v = 12.1 m/s

So this speed is independent of the mass of the rider

5 0
3 years ago
A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance
ArbitrLikvidat [17]

Answer:

The value is v_y  =  -48.61 \ m/s

Explanation:

From the question we are told that

   The horizontal speed is  u_x  = 252 \  m/s

    The horizontal distance is  d = 1250 \ m

Generally the time taken by the hot magma in air before landing is mathematically represented as

       t = \frac{d}{u_x}

=>    t = \frac{ 1250 }{252}

=>    t = 4.96 \  s

Generally the initial vertical velocity of the magma when it was lunched is  

    u_y = 0 \ m/ s

Then the final velocity of the magma when it hits the ground is mathematically represented s

       - v_y  =  u_y + gt

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So  

        - v_y  =  48.61 \ m/s

=>     v_y  =  -48.61 \ m/s

7 0
3 years ago
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