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3 years ago

WHAT DOES DENSITY HAVE TO DO WITH PLATE TECTONICS? Explain

1 answer:

5 0

Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.

Explanation:

Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)

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determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

3 0

3 years ago

A graduated cylinder.measures 15.3 mL. Convert this measurement to DaL

ololo11 [35]

Answer:

0.000153DaL

Explanation:

We have been given:

15.3mL to convert to DaL

DaL is a unit of volume which indicates a decaliter.

This implies that;

1 Da L = 1 x 10²L

So:

1 mL = 1 x 10⁻³L

So 15.3mL will give 15.3 x 10⁻³L

So;

1 x 10²L = 1 DaL

15.3 x 10⁻³L will give $\frac{15.3 x 10^{-3} }{1 x 10^{2} }$ = 15.3 x 10⁻⁵DaL

Therefore, this is 0.000153DaL

5 0

3 years ago

What is TRUE with regard to Bronfenbrenner's social ecological model applied to the context of relationships?

Alisiya [41]

Answer:

Sorry for waisting ur time just tryna get points :)

Explanation:

3 0

3 years ago

Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection.

faltersainse [42]

Answer:

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

Explanation:

We note that the distance traveled by each car after 4 seconds is

Car A = 19 ft in the west direction.

Car B = 26 ft in the north direction

The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.

Therefore, let the distance between the two cars be s

we have

s² = x² + y²

= (19 ft)² + (26 ft)² = 1037 ft²

s = $\sqrt{1037 ft^2}$ = 32.202 ft.

The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

$\frac{ds^{2} }{dt} = \frac{dx^{2} }{dt} + \frac{dy^{2} }{dt}$

→ $2s\frac{ds }{dt} = 2x\frac{dx}{dt} + 2y\frac{dy }{dt}$ which gives

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$

We note that the speeds of the cars were given as

Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

Therefore

$\frac{dy }{dt}$ = 12 ft/sec and

$\frac{dx}{dt}$ = 8 ft/sec

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$ becomes

$32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec} + 26ft\times 12\frac{ft}{sec}$ = 464 ft²/sec

$\frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.}$ = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

3 0

3 years ago

Permafrost thaws only briefly in the summer. t o f brainliest

Verizon [17]

The answer is False. Permafrost does not thaws briefly in the summer since Permafrost is all year round. As global temperature rise, the amount that frozen soil that melts is increasing. With warmer temperature, snow and ice that used to reflect sunlight is disappearing. Thus, the newly exposed soil absorbs heat from the sunlight, and encourages melting.

5 0

3 years ago