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3 years ago

5

If a bat with a mass of 5 kg and acceleration of 2 m/s2 hits a ball whose mass is 0.5 kg in the forward direction, what is the r

eaction force of the ball on the bat?
10 N, forward
0 N, forward
0.5 N, upward
10 N, backward

2 answers:

ivanzaharov [21]3 years ago

8 0

Answer:

10 N , forward

Explanation:

F=m×a

F= 5×2

F=10 N

lina2011 [118]3 years ago

6 0

Answer:

10 N, backward

Explanation:

Force exerted on the ball by the bat can be calculated according to newtons second law,

Force = mass × acceleration

Given mass of the ball = 5kg

Acceleration of the ball = 2m/s²

Force exerted on the ball= 5×2

Force exerted on the ball by the bat = 10N

If the force of 10N is applied on the ball by the bat in the forward direction, the reaction force of the ball on the bat will also be 10N but in the opposite direction (backward) to the direction of the bat according to newton's third law which states that actions and reaction are equal and opposite.

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Water flows with constant speed through a garden hose that goes up to 27.5 cm high. if the water pressure is 132kpa at the botto

sergejj [24]

Answer:

The pressure at the top of the step is 129.303 kilopascals.

Explanation:

From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle ( $v_{bottom}\approx v_{top}$ ):

$p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h$ (1)

$p_{bottom}$ , $p_{top}$ - Total pressures at the bottom and at the top, measured in pascals.

$\rho$ - Density of the water, measured in kilograms per cubic meter.

$\Delta h$ - Height difference of the step, measured in meters.

If we know that $p_{bottom} = 132000\,Pa$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta h = 0.275\,m$ , then the pressure at the top of the step is:

$p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h$

$p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)$

$p_{top} = 129303.075\,Pa$

$p_{top} = 129.303\,kPa$

The pressure at the top of the step is 129.303 kilopascals.

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3 years ago

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

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

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

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

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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)

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