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tankabanditka [31]
3 years ago
10

Striking a fixed object at 60 MPH is the same as falling from a _____ story building:

Physics
2 answers:
alexira [117]3 years ago
7 0
4 story building ............
yulyashka [42]3 years ago
6 0

speed of the striking object is given as

v = 60 mph

now this speed in SI units given as

v = 60 \times \frac{1609 m}{3600 s}

v = 26.82 m/s

So here we can use kinematic now to find the height from where it is dropped

so here we will have

v_f^2 - v_i^2 = 2 a d

now we have

v_f = 26.82 m/s

v_i = 0

a = 9.81 m/s^2

now from above equation

26.82^2 - 0 = 2(9.81)d

d = 36.67 m

now we know that average height of 1 story building is 3 meter

so approx total number of stories in the building will be

N = \frac{36.67}{3} = 12 story

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Do all metals expand the same amount when heated??
ahrayia [7]

Answer:

No

Explanation:

The rate at which solids expand when heated depends on the substance. Metals tend to have higher rates of expansion (per degree change in temperature) than non-metal solids, but there is variation even among metals. A table of expansion coefficients can be found here or here.

8 0
3 years ago
Read 2 more answers
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

8 0
3 years ago
Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c
Sidana [21]

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

7 0
2 years ago
A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string
Darya [45]

Answer:

a

The  speed of  wave is   v_1  = 129.1 \ m/s

b

The new speed of the two waves is v =  129.1 \ m/s

Explanation:

From the question we are told that

    The mass of the string is  m  =  60 \ g  =  60 *10^{-3} \ kg

    The length is  l  =  2.0 \ m

    The tension is  T  = 500 \ N

Now the velocity of the first wave is mathematically represented as

     v_1  = \sqrt{ \frac{T}{\mu} }

Where  \mu is the linear density which is mathematically represented as

      \mu  =  \frac{m}{l}

substituting values    

     \mu  =  \frac{ 60 *10^{-3}}{2.0 }

     \mu  =  0.03\ kg/m

So

   v_1  = \sqrt{ \frac{500}{0.03} }

   v_1  = 129.1 \ m/s

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

     

8 0
3 years ago
Which graphs show the correct relationship between kinetic energy and mass?
Anettt [7]
Kinetic Energy:
Kinetic Energy is the energy of motion.

Mass:
Mass is the number of particles that a substance has


I think the correct graph would be
the THIRD GRAPH
5 0
3 years ago
Read 2 more answers
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