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gtnhenbr [62]
2 years ago
13

Which kind of inclined plane pushes up more? Steeper or flatter?

Physics
1 answer:
Viktor [21]2 years ago
7 0

Answer:

They should've sensed that it was easier as the board got steeper. A flatter inclined plane has to apply more upward force to support something (or someone) because gravity, pulling toward the center of the Earth, pulls the object more into the table.Explanation:

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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
4 years ago
If the battery of your phone can provide 2 mA of current to your phone and holds a charge of 130 C, how long will it take a full
Naya [18.7K]

Answer: 65000 seconds

Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

Charge (Q) = 130 C

Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

Thus, it will take a fully charged phone 65000 seconds to die

5 0
3 years ago
Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i
densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0
3 years ago
During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla
Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of u=-40\ m/s

ball returned with speed of v=40\ m/s

average Force imparted by racquet on the ball is given by

F_{racquet}=\frac{m(v-u)}{\Delta t}

where m=mass\ of\ ball

\Delta t=time of contact of ball with racquet

F_{racquet}=\frac{m(40-(-40))}{\Delta t}

F_{racquet}=\frac{80m}{\Delta t}-----1

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

F_{player}=\frac{m(0-40)}{\Delta t}

F_{player}=\frac{-40m}{\Delta t}-----2

From 1 and 2 we get

F_{racquet}=-2F_{player}

Hence the magnitude of Force by racquet is twice the Force by player

5 0
4 years ago
When a surfer rides an ocean wave on her surfboard, she is actually riding on. A. a crest that is toppling over. . B. a trough o
ruslelena [56]
The right answer to this question is A. a crest that is toppling over. When a surfer rides an ocean wave on her surfboard, she is actually riding on a crest. The crest is the point on a wave with the maximum value or upward displacement within a cycle.
8 0
3 years ago
Read 2 more answers
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