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Pachacha [2.7K]

2 years ago

11

A 0.25 kg baseball is in contact with a bat for 0.0075 s and exerts a force or Tour

1 answer:

Tanzania [10]2 years ago

7 0

Answer:.

Explanation:

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Is it possible for a car to be accelerating to the west while it is driving to the east?

Ber [7]

Answer:

Yes

Explanation:

If the acceleration has an opposite direction to the velocity of the car, this means that it is opposed to is motion. Therefore, it is called deceleration, since the car's velocity will decrease until it stops and then will start it moving towards the west.

8 0

2 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0

3 years ago

How far should the second antenna be moved in order to receive a minimum signal from a station that broadcasts at 99.4 MHz?

raketka [301]

Answer:

So the distance of the antenna from the station will be 3.018 m

Explanation:

We have given the frequency of the broadcast $f=99.4MHz=99.4\times 10^4Hz$

The speed of light $c=3\times 10^8m/sec$

The distance of the antenna to receive a minimum signal from the station is given by $d=\frac{v}{f}=\frac{3\times 10^{8}}{99.4\times 10^6}=3.018m$

So the distance of the antenna from the station will be 3.018 m

5 0

3 years ago

Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims

777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force $F = 12.3$ N

Displacement $d = 10.2$ m

Mass of rocket $m = 0.663$ Kg

From work energy theorem,

$W = \Delta K$

$W_{t} - Wd - W_{g} = KE$

Where $W_{t} =$ thrust work $W_{g} =$ gravitational work

$KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2$

$KE = 59.2 -Wd$

After cutoff kinetic energy is converted into potential energy,

$KE = Wd' + mg\Delta h$

Put value of KE

$59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5$

Work done by drag force is given by,

$Wd'+Wd = 59.2 -29.23$

$= 29.96$ J

Therefore, the work done by the drag force is given by 29.96 J

5 0

2 years ago

Please someone help me.

katovenus [111]

Answer:

acceleration of the car is 3 m\s^2

Explanation:

from rest means the initial velocity (vi) is zero

time = 5s

final velocity (vf) = 15m\s

a = vf - vi \ t

a = (15-0) \ 5

a= 3 m\s^2

which means that the car is speeding up 3 meters every second

5 0

3 years ago