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madam [21]
2 years ago
7

A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 20 m/s. Assuming

air resistance is negligible, find the speed (magnitude of velocity) in meters per second of the shell at launch and 4.7 s after the launch.
Physics
1 answer:
ivolga24 [154]2 years ago
3 0

Answer: The launch speed is 43m/s

4.7secs after launch speed is 4m/s

Explanation: To solve this we use the first equation of motion but in this case our acceleration would be -10m/s² since we are going upwards against gravity(launch).

Vf = Vi - a*t

Where Vf is the final velocity after launch, Vi is the initial velocity at launch, t is time in secs then a is acceleration

a. From the question

t = 2.3secs

Vf = 20m/s

a = -10m/s²

Substituting into the above equation we have that,

20= Vi - 10* 2.3

20 = Vi - 23

Vi = 20+23

Vi = 43 m/s

Which is the speed at launch.

b. The magnitude of speed (Vf) 4.7 sec after launch is calculated as follows using same procedure but here Vi is 43m/s as calculated

Vf = 43 - 10*4.7

Vf = 43 - 47

Vf = -4m/s

But since we are asked to find the magnitude we neglect the negative sign.

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The speed of light is greater in a vacuum than in air or water.<br><br> True or false
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A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the
krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒  Ycm = (ms*D + mp*0) / (ms + mp)

⇒  Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

5 0
2 years ago
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon
Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

F_{branch} + F_{gravity}=F_{centripetal}

force of gravity = - mg

F_{branch}=F_{centripetal}-F_{gravity}

                        = \dfrac{mv^2}{r} + mg

                        = m(\dfrac{3.4^2}{0.6} +9.8)

                        =9 x 29.067

                        = 261.6 N

upward force acting = 261.6 N

7 0
2 years ago
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