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seropon [69]
3 years ago
5

What is 8 m/s - 3 m/s?

Physics
1 answer:
Dima020 [189]3 years ago
3 0

5m/s 8-3 is five so five


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A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt
anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

6 0
3 years ago
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a)
Marysya12 [62]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is A_{x} = 0

the y-component of the force at A is A_{y}  = 400 N

the couple acting at A is; M_{A} = 146 N-m

b)

the sum of the momentum about the right end of the beam is;  ∑M_{R}  = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑M_{L} = 0

Vector force acting at A, F_{A} = A_{x}i + A_{y}j

Now, From the image, we have;

a)

∑f = 0

F_{A} - 600j + 200j = 0i + 0j

A_{x}i + A_{y}j - 600j + 200j = 0i + 0j

A_{x}i + (A_{y} - 400)j = 0i + 0j

now by equating i- coefficients'

A_{x} = 0

so, the x-component of the force at A is A_{x} = 0

also by equating j-coefficient

A_{y} - 400 = 0

A_{y}  = 400 N

hence, the y-component of the force at A is A_{y}  = 400 N

we also have;

∑M_{L} = 0

M_{A}  - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

M_{A} - 30 N-m - 228 N-m + 112 Nm = 0

M_{A} - 146 N-m = 0

M_{A} = 146 N-m

Therefore, the couple acting at A is; M_{A} = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑M_{R} = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)(A_{y} ) + M_{A}

∑M_{R} = (108  N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑M_{R} = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑M_{R}  = 0

Therefore, the sum of the momentum about the right end of the beam is;  ∑M_{R}  = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

7 0
3 years ago
Physics Question:<br> (ignore the part crossed in red) 14459688
adell [148]
Spinning a marshmallow over a fire is effective maybe if you hang it over the fire and heat it up equally on each side
3 0
3 years ago
What us the formula of finding diameter
Over [174]

Answer:

The formula to find the diameter states the relationship between the diameter and the radius. The diameter is made up of two segments that are each a radius. Therefore, the formula is: Diameter = 2 * the measurement of the radius. You can abbreviate this formula as d=2r.

Explanation:

5 0
3 years ago
How does the wavelength of a wave change when frequency decreases? when frequency increases?
Law Incorporation [45]

When the frequency decreases the wavelength is further apart. When it increases its closer together. Think about a flat line when the frequency is low the wavelengths are wider. When its a high frequency the squiggly lines on the moniter are taller and thinner so the wavelengths are not as wide and not that far from each other depending on how high the frequency is.

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