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d1i1m1o1n [39]
1 year ago
11

The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me

ters when t is in seconds. What will be the dimensions of b and c in this case if u = 8 and v = 7?
Physics
1 answer:
KIM [24]1 year ago
8 0

As per given equation we have

x = bt^u + ct^v

now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation

now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

bt^u = M^0L^1T^0

b*T^8 = M^0L^1T^0

b = M^0L^1T^{-8}

similarly for other term we have

ct^v = M^0L^1T^0

c*T^7 = M^0L^1T^0

c = M^0L^1T^{-7}

<em>so above are the dimensions of b and c</em>

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(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind
Vilka [71]
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
F_f = \mu N
where \mu is the coefficient of friction, while N is the normal force. So we have:
F=\mu N
since we know that F=300 N and \mu=0.3, we can find N, the magnitude of the normal force:
N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is \mu=0.03 due to the presence of the oil. Therefore, we have:
N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N
8 0
2 years ago
A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc
Tasya [4]

Answer:

17 m/s south

Explanation:

m_1 = Mass of dog = 10 kg

m_2 = Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

u_1 = Velocity of dog = 1 m/s

u_2 = Velocity of skateboard

In this system the linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

The velocity of the skateboard will be 17 m/s south as the north is taken as positive

3 0
2 years ago
A 65 kg students is walking on a slackline, a length of webbing stretched between two trees. the line stretches and sags so that
polet [3.4K]

Answer : Tension in the line = 936.7 N

Explanation :

It is given that,

Mass of student, m = 65 kg

The angle between slackline and horizontal, \theta=20^0

The two forces that acts are :

(i) Tension

(ii) Weight

So, from the figure it is clear that :

mg=2T\ sin\ \theta

T=\dfrac{mg}{2\ sin\theta}

T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}

T=936.7\ N

Hence, this is the required solution.

5 0
1 year ago
Read 2 more answers
Which of these is true of a chemical reaction? (a) energy is always transferred from reactants to products. (b) Energy is transf
lorasvet [3.4K]
The answer is B: energy is transferred, but it can go to the products or the reactants.
3 0
2 years ago
Read 2 more answers
A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
yanalaym [24]

Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

That would be our first equation

Now, we need to analyze its vertical motion

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Knowing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Solving for t

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

So, the magnitude of the velocity is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The <u><em>direction of the ball</em></u> is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

4 0
2 years ago
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