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brilliants [131]
3 years ago
7

Which statements about the two fossil images are correct?

Physics
2 answers:
tangare [24]3 years ago
5 0
What fossil images, there are no pictures attached.
Ostrovityanka [42]3 years ago
4 0

the answer is C and D on plato

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3. what could you do to increase the precision of your measurements? What factors limit the
Vika [28.1K]

Answer:

Use specific tools built specifically for that specific measurement.

Explanation:

8 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
2 years ago
Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha
Ksivusya [100]

Answer:

A_c=87.73*10^{21}m/s

Explanation:

From the question we are told that

r=5\times 10^{-11}

T=1.5 \times 10^{-16}

Generally the equation for velocity is mathematically given as

Velocity (v)=\frac{2 \pi r}{t}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

Generally the equation for Centripetal acceleration is mathematically given as

A_c=\frac{V^2}{r}

A_c=(\frac{20.944*10^5)}{r5*10^{-11}}

A_c=87.73*10^{21}m/s

8 0
2 years ago
Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro
Natasha_Volkova [10]

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

F \  \alpha \  \frac{1}{D^{2}} for a distance D

If we move them so that D is doubled:

\frac{1}{2^{2}.D^{2}  }= \frac{1}{4} \eq  \frac{1}{.D^{2}  } \eq

Then the force they experience is one fourth of the original.

5 0
3 years ago
A 40kg rock falls off a cliff that is 50 meters high. How fast is the speed of the rock when it hits the ground
Oksi-84 [34.3K]

31.3m/s

Explanation:

Given parameters:

Mass of rock = 40kg

Height of cliff = 50m

Unknown:

Speed of rock when it hits ground = ?

Solution:

We are going to use the appropriate motion equation to solve this problem

The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.

  Using;

                      V²  = U² + 2gH

V = unknown velocity

U = initial velocity = O

g = acceleration due to gravity = 9.8m/s²

H = height of fall

since the initial velocity of the bodyg is 0

     V²  = 2gH

    V= √2gH = √2 x 9.8 x 50 = 31.3m/s

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

7 0
3 years ago
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