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iris [78.8K]
3 years ago
15

A(n) ______ is a gap in the geologic record where some rock layers have been lot because of erosion

Physics
1 answer:
fomenos3 years ago
6 0

geological time is the answer

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klemol [59]
Diagram B .... light shines through at an angle
6 0
3 years ago
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What is cheetah speed and strength?
pav-90 [236]

Answer:

Speed: 109.4–120.7 km/h (68.0–75.0 mph)

Strength: Couldn't find out.

Explanation:

3 0
2 years ago
Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s .
lara [203]

Answer:

3.467 s

Explanation:

given,

distance , d = 49 mm = 0.049 m

initial speed of the of the rock, v = 17 m/s

time taken by the Heather rock to reach water

using equation of motion

s = ut +\dfrac{1}{2}at^2

taking downward as negative

-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2

4.9 t² + 17 t - 0.049 = 0

now,

t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}

t₁ = -3.47 s , 0.0028 s

rejecting negative values

t₁ = 0.0028 s

now, time taken by the ball of Jerry

using equation of motion

s = ut +\dfrac{1}{2}at^2

taking downward as negative

-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2

4.9 t² - 17 t - 0.049 = 0

now,

t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}

t₂ = 3.47 s ,-0.0028 s

rejecting negative values

t₂ = 3.47 s

now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s

5 0
3 years ago
What are some everyday products we use made from refined metal or ore
FinnZ [79.3K]

Answer:

Hand tools.

Bolts, nuts, and screws.

Cans.

Cutlery.

Pipes and pipe fittings.

Metal windows and doors.

Equipment attachments.

Car parts.

Explanation:

4 0
3 years ago
A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface
Fiesta28 [93]

Answer:

X_2=25.27m

Explanation:

Here we will call:

1. E_1: The energy when the first spring is compress

2. E_2: The energy after the mass is liberated by the spring

3. E_3: The energy before the second string catch the mass

4. E_4: The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. E_1 = E_2

2. E_2 -E_3= W_f

3.E_3 = E_4

where W_f is the work of the friction.

1. equation 1 is equal to:

\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2

where K is the constant of the spring, x is the distance compressed, M is the mass and V_2 the velocity, so:

\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2

Solving for velocity, we get:

V_2 = 65.319 m/s

2. Now, equation 2 is equal to:

\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd

where M is the mass, V_2 the velocity in the situation 2, V_3 is the velocity in the situation 3, U_k is the coefficient of the friction, N the normal force and d the distance, so:

\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)

Volving for V_3, we get:

V_3 = 65.27 m/s

3. Finally, equation 3 is equal to:

\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2

where K_2 is the constant of the second spring and X_2 is the compress of the second spring, so:

\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2

solving for X_2, we get:

X_2=25.27m

3 0
3 years ago
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