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2 years ago

Using your strategy to overcome your obstacle.

Physics

2 answers:

Rama09 [41]2 years ago

7 0

Answer:

Don’t complain. People don’t want to hear woe is me over and over again, especially if you do nothing about it. However, do ask for help and for suggestions from others that may have been in similar situations. You have to be willing to help yourself. People can’t do it for you.

Explanation:

USPshnik [31]2 years ago

6 0

Answer:

buena suerte con las citas del futuro

Explanation:

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What centripetal force is needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s

liq [111]

Answer:

We conclude that the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s is 393.75 N.

Explanation:

Given

Mass m = 7 kg

Radius r = 4 m

Velocity v = 15m/s

To determine

We need to determine the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s.

We know a centripetal force acts on a body to keep it moving along a curved path.

We can determine the centripetal force using the formula

$F_c=\frac{mv^2}{r}$

where

$m$ is the mass

$v$ is the velocity

$r$ is the radius

$F_c$ is the centripetal force

substitute m = 7, r = 4, and v = 15 in the formula

$F_c=\frac{mv^2}{r}$

$F_c=\frac{7\left(15\right)^2}{4}$

$F_c=\frac{1575}{4}$

$F_c=393.75$ N

Therefore, we conclude that the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s is 393.75 N.

7 0

3 years ago

A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a

Readme [11.4K]

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

$H = 4.905 m$

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

$a = g = 9.81 m/s^2$

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

$v_f = 0$

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

$\Delta y = 0 = v_y t + \frac{1}{2} at^2$

$0 = v_y (2) - \frac{1}{2}(9.81)(2^2)$

$v_y = 9.81 m/s$

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

here we have

$0 - (9.81^2) = 2(-9.81)H$

$H = 4.905 m$

4 0

3 years ago

What advantage do weather do weather satellites have over ground based weather stations

docker41 [41]

Answer:

Satellites can gather weather data from much higher altitudes than land-based instruments can.

3 0

3 years ago

What is the magnitude of the centripetal force that must be applied in order for a 2kg ball on a 2.0 m string to spin with unifo

NeTakaya

The ball needs to accelerate with a magnitude a (pointed towards the center of the circle) of

a = (5.0 m/s)² / (2.0 m) = 12.5 m/s²

Then the required tension F in the string would need to be

F = (2 kg) (12.5 m/s²) = 25 N

6 0

2 years ago

Read 2 more answers

A particle moves in a straight line so that time t seconds its displacement is x metres from a fixed point is given by x=3t^2 +2

Andru [333]

Answer:

-5 meters

Explanation:

x = 3t² + 2t − 5

The initial displacement is when t = 0.

x = 3(0)² + 2(0) − 5

x = -5

6 0

3 years ago