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Aleksandr [31]
3 years ago
8

A scientist is conducting research to determine how much pressure a dam can withstand. Why are good observations important to hi

s research?
Physics
1 answer:
Natalija [7]3 years ago
3 0

The things that a scientist should consider while observing the force is the environmental conditions,the force that is expected to act on the dam, the means to contain that force,  and compare different types of designs in accordance with the location of the dam

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At an amusement park, a roller coaster carrying nine homosexuals, moving at 18 m/s, jumps its track at a point that is 20 meters
Lorico [155]

It is the same if a white father and a white mother give birth to a black kid, so the mother accuses the father for having (let's say mating) with a black women.

6 0
3 years ago
At an accident scene on a level road, investigators measure a car’s skid mark to be 88 m long. The accident occurred on a rainy
Oduvanchick [21]

Answer:

The the speed of the car is 26.91 m/s.

Explanation:

Given that,

distance d = 88 m

Kinetic friction = 0.42

We need to calculate the the speed of the car

Using  the work-energy principle

work done = change in kinetic energy

W=\Delta K.E

\mu\ mg\times d=\dfrac{1}{2}mv^2

v^2=2\mu g d

Put the value into the formula

v=\sqrt{2\times0.42\times9.8\times88}

v=26.91\ m/s

Hence, The the speed of the car is 26.91 m/s.

3 0
3 years ago
Read 2 more answers
The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
3 years ago
Can concave lens be used to make hand lens .why?​
mario62 [17]

Answer:

No concave lens cannot be used to make hand lens because in hand Lens we use convex lens so as to converge the rays of the light After refraction and helps to produce a magnified image of an object.

Hope it will help you :)❤

8 0
3 years ago
A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is
goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, v_i = 1.4 m/s

final velocity of the child, v_f = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, f_k = 41 N

The work done by the child is calculated as;

\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech}  + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96  \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0
3 years ago
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