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1 year ago

15

What is the net force in this image?

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