A cheetah starts from rest and accelerated at 8.7 m/s^2 for 3s. How far did the cheetah go in that time

nexus9112 [7]

C i’m pretty sure. correct me if i’m wrong :)

7 0

3 years ago

A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the

o-na [289]

Answer:

A. As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position.

Gpe @ A=mgh=40*9.81*2=784.8J

B. At 30degree vertical angle the vertical displacement from lowest position is given by

2-2cos(30)=2-1.73=0.27m

Gpe @B= 40*9.81*0.27=106 J

C: at the bottom of circular arc it's Gpe is zero relative to lowest position as bottom of arc itself is lowest position.

8 0

3 years ago

A telephone wire has a current 20A flowing through it . How long does it take for a charge of 15 C to pass through the wire?

Sauron [17]

Charge = Current/time
Q = It
t = Q/I = 15/20 = 0.75 s

7 0

4 years ago

A man is trying to push a 250 N dresser across his carpeted bedroom. He applies a force of 20 N and the carpet provides a fricti

11111nata11111 [884]

Answer:

The correct option is;

B. Subtract vectors A from C; $F_{Net}$ = 12 N

Explanation:

The given parameters are;

The weight of the dresser = 250 N

The force applied by the man = 20 N = C

The frictional force provided by the carpet = 8 N = The component of the weight resisting motion = A

Therefore, the net force tending to put the dresser in motion, $F_{Net}$ , is given as follows;

$F_{Net}$ = The force applied by the man, C - The frictional force provided by the carpet which is the component of the weight resisting motion, A

$F_{Net}$ = C - A = 20 N - 8 N = 12 N

$F_{Net}$ = 12 N

The resulting net force, $F_{Net}$ = 12 N.

3 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

2 years ago

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