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tamaranim1 [39]
3 years ago
10

If a baseball is in space and someone hits it with a bat what does the ball do

Physics
1 answer:
Darina [25.2K]3 years ago
3 0

Answer:

It's will fly like float

Explanation:

Since in space there is no gravity that everything can float

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A
ale4655 [162]

m = 43.2 kg

Explanation:

volume of sphere = (4/3)pi(r)^3

= (4/3)(3.14)(2 m)^3

= 33.5 m^3

density = mass/volume

or solving for mass m,

m = (density)×(volume)

= (1.29 kg/m^3)(33.5 m^3)

= 43.2 kg

3 0
3 years ago
A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th
forsale [732]

Answer:

F=-18412.9N, where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

1\ hour = 3600\ seconds

1\ mile = 1600\ meters

1000g = 1kg

We can write that:

\frac{1\ hour}{3600\ seconds}=1

\frac{1600\ meters}{1\ mile}=1

\frac{1kg}{1000g}=1

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s

110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s

145 g=145 g(\frac{1kg}{1000g})=0.145kg

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is a=\frac{\Delta v}{\Delta t}, so we have:

F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}

Taking the throw direction as the positive one, for our values we have:

F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N

4 0
3 years ago
Which type of lens is shown in the picture below?<br> plane<br> refractional<br> concave
TiliK225 [7]
It is a concave lens

Have a nice day
8 0
3 years ago
What is the distance a truck can travel if it moves 7 m/s for 20 seconds?
artcher [175]

if it moves 7m/s, that means every second it goes 7m.  Now we just multiply by the time (20 seconds) and end up with 140m.

6 0
3 years ago
Read 2 more answers
A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge
Troyanec [42]

Answer:

I_{l} =44.84 \mu A

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire =  1.30×10⁻² Ω  

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))

\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))

I_{l} = \dfrac{V}{R}

I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}

I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))

I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))

I_{l} =44.84 \times 10^{-6}\A

I_{l} =44.84 \mu A

3 0
3 years ago
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