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quester [9]
3 years ago
13

In subduction, _____.

Physics
1 answer:
user100 [1]3 years ago
4 0

The answer is the less dense plate slides over the denser plate.

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An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a
vladimir1956 [14]
I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s
8 0
3 years ago
Read 2 more answers
Definition of the following word <br>Work ,measurement​
garri49 [273]

Answer:

measurements

Explanation:

Measurement is the assignment of a number to a characteristic of an object or event, which can be compared with other objects or events. The scope and application of measurement are dependent on the context and discipline.

7 0
4 years ago
Read 2 more answers
(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en
V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0
3 years ago
The period T of a pendulum of length L is measured to determine g at the surface of Earth. The equation used is T=2π√L/g. The ma
saul85 [17]

Answer:

C: Variation in the value of g as the pendulum bob moves along its arc.

Explanation:

The formula for period of a simple pendulum is given by;

T = 2π√(L/g)

Where;

L is length

g is acceleration due to gravity

Now, from this period equation, it is clear that the only thing that can affect the period of a simple pendulum are changes to its length and acceleration due to gravity.

Looking at the options, the only one that talks about either the length or gravity as being potential causes of the error is option C

4 0
3 years ago
A fireworks shell is accelerated from rest to a velocity of 68.0 m/s over a distance of 0.230 m.
Ugo [173]

Answer:

(A) 10052.2 m/s²

(B)  0.00678 seconds

Explanation:

From the question,

(A) Applying

V² = U²+2as..................... Equation 1

Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.

make a the subject of the  equation

a = (V²-U²)/2s........................ Equation 2

Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m

Substitute these values into equation 2

a = (68²-0²)/(2×0.230)

a = 10052.2 m/s²

(B) Using,

a = (V-U)/t......................... Equation 3

Where t= time.

make t the subject of the equation

t = (V-U)/a......................... Equation 4

Given: V = 68 m/s, U = 0 m/s, a = 10052.2

Substitute into equation 4

t = (68-0)/10052.2

t = 0.00678 seconds

5 0
4 years ago
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