A baseball player hits a ball in such a way that it leaves bat with an initial speed of 40.0 m/s at 38.0 degrees.

Earth turns on its axis about once every 24 hours. The Earth's equatorial radius is 6.38 x 106 m. If some catastrophe caused Ear

san4es73 [151]

To solve this problem it is necessary to apply the concepts related to the Period of a body and the relationship between angular velocity and linear velocity.

The angular velocity as a function of the period is described as

$\omega = \frac{2\pi}{T}$

Where,

$\omega =$ Angular velocity

T = Period

At the same time the relationship between Angular velocity and linear velocity is described by the equation.

$v = \omega r$

Where,

r = Radius

Our values are given as,

$T = 24 hours$

$T = 24hours (\frac{3600s}{1 hour})$

$T = 86400s$

We also know that the radius of the earth (r) is approximately

$6.38*10^6m$

Usando la ecuación de la velocidad angular entonces tenemos que

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{86400}$

$\omega = 7.272*10^{-5}rad/s$

Then the linear velocity would be,

$v = \omega *r$

x $v = \omega *r$

$v= 463.96m/s$

The speed would Earth's inhabitants who live at the equator go flying off Earth's surface is 463.96

3 0

3 years ago

A police officer measures a straight skid mark on the road of length d=20m. the car involved is stationary at the end of the ski

zvonat [6]

80 divided by 20 is 2

8 0

3 years ago

A ball rolls 60 centimeters along a sidewalk in 5 seconds what is the speed of the ball.

meriva

Answer:

12 cm/s

Explanation:

Quite simply, you are looking for cm/s

so 60 cm / 5s = 12 cm/s

6 0

2 years ago

Read 2 more answers

A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel

anzhelika [568]

Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck $a_{t$ = 1 m/ $s^{2}$ (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

$a_{A}$ = 0.5 m/ $s^{2}$ (upward) , $a_{B}$ = 0.5 m/ $s^{2}$ (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( $m_{A}$ + $m_{B}$ )g = ( $m_{A}$ + $m_{B}$ ) $a_{B}$

= 2T- (400 + 50)*(9.81 m/ $s^{2}$ ) = (400 + 50)*(0.5 m/ $s^{2}$ )

T = 2320N

Truck: $M_{R}$ = ∑( $M_{R}$ )eff: = $-N_{f}$ (3.4m) + $m_{T}$ (2.0m) - T (0.6m)= $m_{T} a_{T}$ (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/ $s^{2}$ )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/ $s^{2}$ ) = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: $N_{f}$ + $N_{R}$ - $m_{T}$ g = 0

10544 + $N_{R}$ - (2000kg)(9.81 m/ $s^{2}$ ) = 0

$N_{R}$ = 9076N

∑fx (to the left) = ∑(fx)eff: $F_{R}$ - T = $m_{T} a_{T}$

$F_{R}$ = 2320N + (2000kg)(9.81 m/ $s^{2}$ ) = 4320N

(a) reaction at each front wheel:

1/2 $N_{f}$ (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: $N_{B}$ + $M_{B}$ g - $m_{B}$ $a_{B}$

$N_{B}$ = (400kg)(9.81 m/ $s^{2}$ ) + (400kg)(0.5 m/ $s^{2}$ ) = 4124N (compression)

3 0

4 years ago

Read 2 more answers

30 points!

vlada-n [284]

Answer:

Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.

Again item 2 refers to bonds between atoms and is eliminated.

This leaves only item 1.

Nuclear forces are very short range forces between components of the nucleus.

Weak nuclear forces are trillions of times smaller than strong forces.

Gravitational forces are much much smaller than the weak nuclear force.

6 0

3 years ago