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2 years ago

8

Show that the units N/kg can be written using only units of meters (m) and second (s). Is this a unit of mass ,acceleration or f

orce.

1 answer:

nikdorinn [45]2 years ago

3 0

<span>When the difference between two results is larger than the estimates error, the result is</span>

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Which one of the following statements concerning superconductors is false? A constant current can be maintained in a superconduc

julsineya [31]

Answer:

All materials are superconducting at temperatures near absolute zero kelvin.

Explanation:

All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.

8 0

3 years ago

The area of a rectangular park is 4 mi^2. The park has a width that is equal to "w", and a length that is 3 mi longer than the w

goldfiish [28.3K]

Answer:

l= 4 mi : width of the park

w= 1 mi : length of the park

Explanation:

Formula to find the area of the rectangle:

A= w*l Formula(1)

Where,

A is the area of the rectangle in mi²

w is the width of the rectangle in mi

l is the width of the rectangle in mi

Known data

A = 4 mi²

l = (w+3)mi Equation (1)

Problem development

We replace the data in the formula (1)

A= w*l

4 = w* (w+3)

4= w²+3w

w²+3w-4= 0

We factor the equation:

We look for two numbers whose sum is 3 and whose multiplication is -4

(w-1)(w+4) = 0 Equation (2)

The values of w for which the equation (2) is zero are:

w = 1 and w = -4

We take the positive value w = 1 because w is a dimension and cannot be negative.

w = 1 mi :width of the park

We replace w = 1 mi in the equation (1) to calculate the length of the park:

l= (w+3) mi

l= ( 1+3) mi

l= 4 mi

8 0

3 years ago

You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi

Elenna [48]

Answer:

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

Explanation:

The period of the simple pendulum is:

$T = 2\pi\cdot \sqrt{\frac{l}{g} }$

Where:

$l$ - Cord length, in m.

$g$ - Gravity constant, in $\frac{m}{s^{2}}$ .

Given that the same pendulum is test on each planet, the following relation is formed:

$T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}$

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

$\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

5 0

3 years ago

Needddd helppppppp!!!

yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

4 0

2 years ago

Thought Experiment: A monkey escapes from a zoo and climbs a tree. After failing to entice the monkey down, a zookeeper fires a

Andreyy89

Explanation:

When bullet is shot towards the monkey then let say the distance of monkey from the bullet is "d"

so we can find the time to reach the bullet to the monkey

$t = \frac{d}{vcos\theta}$

Now similarly we can find the vertical displacement of the bullet in the same time

$\Delta y = vsin\theta t - \frac{1}{2}gt^2$

$\Delta y = v sin\theta (\frac{d}{vcos\theta}) - \frac{1}{2}gt^2$

so it is given as

$\Delta y = d tan\theta - \frac{1}{2}gt^2$

here if the monkey is initially at height H above the ground at given angle then we can say

$H = dtan\theta$

so we can say that

$\Delta y = H - \frac{1}{2}gt^2$

So if at the same time monkey will fall down then the height of monkey from ground after time "t" is given as

$\Delta y = H - \frac{1}{2}gt^2$

so here bullet will hit the monkey as both monkey and bullet are at same position.

3 0

2 years ago