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siniylev [52]
2 years ago
13

"the velocity of the block is constant, so the net force acting on the block must be zero. Thus the normal force equals the weig

ht and the force of friction equals the applied force" what, if anything, is wrong with this statement
Physics
1 answer:
RSB [31]2 years ago
6 0

Answer:

Explanation:

"the velocity of the block is constant, so the net force acting on the block must be zero. Thus the normal force equals the weight and the force of friction equals the applied force" The only thing which is wrongly stated here is that the normal force equals the weight. Normal force is not always equal to weight. If external force is applied at some angle with the horizon , normal force may be more or less than weight. Even in case of an object placed on an  inclined plane , weight is not equal to normal force .

All other parts of the statement are correct.

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If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
3 years ago
Which term below describes a measurement of how hard an object pushes against a surface?
Scilla [17]

I hope the answer is D. Pressure

8 0
2 years ago
Read 2 more answers
A balance uses small metal weights of known mass to determine the mass of a substance. Where would a balance NOT function correc
tatuchka [14]

Answer:

D is the answer.

Explanation:

Just do it.

6 0
3 years ago
Read 2 more answers
2)
earnstyle [38]

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

x=0.150m-0.115m=0.035m

And spring constant is k=51.0N/m

PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312

Hence, the potential energy is 0.0312J

8 0
3 years ago
The displacement of a wave traveling in the negative y-direction is D(y,t) = ( 4.60 cm ) sin ( 6.20 y+ 60.0 t ), where y is in m
trapecia [35]

Answer:

The question is incomplete, below is the complete question

"The displacement of a wave traveling in the negative y-direction is D(y,t) = ( 4.60cm ) sin ( 6.20 y+ 60.0 t ), where y is in m and t is in s.

A) What is the frequency of this wave?

B)  What is the wavelength of this wave?

C) What is the speed of this wave?"

Answers:

a.  f=\frac{30}{\pi }Hz\\

b. wavelength=\frac{\pi }{3.1}m \\

c. v=9.68m/s

Explanation:

The equation of a wave is represented as

D(x,t)=Asin(kx+wt) \\

Where A=amplitude

w=angular frequency=2πf

K=wave numbers =2π/λ

since we re giving he equation  D(y,t) = ( 4.60cm ) sin ( 6.20 y+ 60.0 t ),

we can compare and get the value for the wave number and angular frequency.

By comparing we have

w=60rads/s

k=6.20

a. to determine the frequency, from the expression fr angular wave frequency we have

w=2πf hence

f=w/2π

if we substitute we arrive at

f=\frac{60}{2\pi }\\f=\frac{30}{\pi }Hz\\

b. to determine the wave length, we use

k=\frac{2\pi }{wavelength} \\k=6.2\\wavelength=\frac{2\pi }{k} \\wavelength=\frac{2\pi }{6.2} \\wavelength=\frac{\pi }{3.1}m \\

c. the wave speed  v is express as the product of the frequency and the wavelength. Hence

v=frequency*wavelength \\v=\frac{30}{\pi } *\frac{\pi }{3.1}\\ v=9.68m/s

6 0
3 years ago
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