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GrogVix [38]
4 years ago
9

How fast must a 350 gram softball be traveling in order to have a momentum of .600 kg

Physics
1 answer:
BartSMP [9]4 years ago
4 0
Momentum is mass times velocity. So here we can just substitute in our givens and solve for velocity.

.600kg*m/s/.350kg=1.71m/s

Hope this helps! Thank you!
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Determine the velocity (in m/s) of the object at 10seconds. Include a numerical answer accurate to the first decimal place. If n
tatuchka [14]

Answer:

v = 0.5 [m/s]

Explanation:

In order to determine the speed in such a time interval, we must calculate the slope between the last two positions.

The slope of a line is determined by the following mathematical expression.

P₂ = point 2 = (12,12) = (x₂,y₂)

P₁ = Point 1 = (6,9) = (x₁,y₁)

In this specific case, we must see that in the x-axis we have time, and on the y-axis, we have the space axis.

Now using the slope:

m =\frac{y_{2} -y_{1} }{x_{2} -x_{1} } \\m = \frac{12-9}{12-6}\\m = 0.5

This slope is equal to the speed (velocity)

v = 0.5 [m/s]

8 0
3 years ago
A proton is accelerated down a uniform electric field of 450 N/C. Calculate the acceleration of this proton.
olasank [31]

Answer:

The acceleration of proton will be 431.137\times 10^8m/sec^2

Explanation:

We have given electric field E = 450 N/C

Charge on proton =1.6\times 10^{-19}C

Force on electron due to electric field is given by F=qE=1.6\times 10^{-19}\times 450=720\times 10^{-19}N

Mass of electron m=1.67\times 10^{-27}kg

Now according to second law of motion F=ma

So 720\times10^{-19}=1.67\times 10^{-27}a

a=431.137\times 10^8m/sec^2

So the acceleration of proton will be 431.137\times 10^8m/sec^2

6 0
3 years ago
A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
Lilit [14]

Answer:

v = 1.08 m/s

Explanation:

What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

The decrease in PE is

d = 80.0cm * 1 / 1000m = 0.80m

h = 0.80 m /2 = 0.40 m

ΔPE = m*g*h

ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m

ΔPE = 0.1176 J

The moment of inertia of the assembly is

I = 1/12*m*L² + (m1 + m2)*(L/2)²

I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²

I = 0.032 kg·m²

KE = ½Iω²

0.1176 J = ½ * 0.032kg·m² * ω²

ω = 2.71 rad/s

v = ωr = 2.71 rad/s * 0.400m

The linear velocity

v = 1.08 m/s

3 0
3 years ago
An isolated system consists of two masses. The first, m1, has a mass of 1.90 kg, and is initially traveling to the east with a s
Natali [406]

Answer:

m1v1=m2v2, v2=4.3m/s KE=(0.5)(2.94)(4.3)=6.2J

3 0
3 years ago
A plane flies along a straight line path after taking off, and it ends up 210 km farther east and 90.0 km farther north, relativ
Lapatulllka [165]
<span>A plane flies along a straight line path after taking off, and it ends up 210 km farther east and 90.0 km farther north, relative to where it started. we should use Pythagoras theorem for this. distance^2= 210^2 + 90^2.=228.47km, when rounding off this would comes around 230 km</span>
8 0
3 years ago
Read 2 more answers
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