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3 years ago

3.

1 answer:

5 0

A. Acceleration= change in velocity divided by the change in time so
6 m/s/1.5 s = 4 m/(s^2)

B. Speed x time= 6 x 1.5 = 9 m

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HELP PLS! :/

zlopas [31]

Explanation:

formula: Vi = Vf - (at)

Vi: intial velocity

Vf: final velocity

a: acceleration

t: time

fill in formula with the numbers you are given

Vi= 41.6m - ((9.81 m/s^2)(1.89s))

parenthesis first according to pemdas

Vi= 41.6m - 18.54m/s

Answer: 23.06m/s

disclaimer: I havent done physics in awhile so I have no idea if this is right. just an attempt to help steer you in the right direction hopefully. good luck

7 0

3 years ago

If you are riding on a bus with a friend, you can tell you are<br> moving by observing the motion of

nikklg [1K]

Other object outside the bud

8 0

3 years ago

A projectile's horizontal range on level ground is r=v20sin2θ/g. at what launch angle or angles will the projectile land at half

Ksju [112]

As given the formula of range is

$R = \frac{v^2sin2\theta}{g}$

now for the maximum range

$\theta = 45$

$R_{max} = \frac{v^2}{g}$

now for half of this maximum range we will have

$\frac{R_{max}}{2} = \frac{v^2sin2\theta}{g}$

$\frac{v^2}{2g} = \frac{v^2sin2\theta}{g}$

now from above we have

$sin2\theta = \frac{1}{2}$

so two possible values for above is given as

$\theta = 15 degree, 75 degree$

so above two angle we will have half of maximum range

6 0

4 years ago

Consider an eraser, weighing 20g, sitting 5cm from the center of a turntable (flat disk attached to a motor which can spin the d

Artist 52 [7]

Answer:

A. 42.3 revolution per minute

B. Centripetal Force

Explanation:

20 g = 0.02 kg

5 cm = 0.05 m

Let g = 9.8 m/s2

A.The friction force acting on the eraser is the product of normal force and its coefficient:

$F_f = N\mu = mg\mu = 0.02*9.8*0.1 = 0.0196 N$

For the eraser to begin slipping, its centripetal acceleration must win over the acceleration created by friction force, which is

$a_f = F_f / m = 0.0196 / 0.02 = 0.98 m/s^2 = a_c$

We can calculate the angular velocity from this:

$a_c = \omega^2 r$

where r = 0.05 is the radius of rotation, which is distance from the center to the eraser

$\omega^2 = a_c/r = 0.98 / 0.05 = 19.6$

$\omega = \sqrt{19.6} = 4.43 rad/s$

If it spins and angle of 4.43 rad per second then for every minute (60 seconds) it would spin an angle of 4.43 * 60 = 265.6 rad/min

Since 1 revolution is 2π rad, then in a minute it would spin 265.6 / 2π = 42.3 revolution/minute

B. The force responsible for making the eraser go faster while the turntable is spinning up would be the centripetal force.

7 0

3 years ago

How is an experiment is actually conducted

NeX [460]

By using the scientific method

3 0

3 years ago