If a box that weights one pound (5 newtons) needs to be moved, how much force needs to be applied to move it?

The small piston of a hydraulic lift, has an area of 0.01m2. If a force of 250N is applied to the small piston if it has an area

postnew [5]

Answer:

Explanation:

Given that,

Small piston Hydraulic piston has an area

A1 = 0.01m²

If the force applied is 250N is applied to the small piston at an area of 0.05 m²

Then,

F2 = 250 N and A2 = 0.05m²

Then, applying pascal principle,

Pressure at small area = pressure are bigger area

P1 = P2

F1 / A1 = F2 / A2

F1 / 0.01 = 250 / 0.05

F1 / 0.01 = 5000

Cross multiply

F1 = 5000 × 0.01

F1 = 50 N

4 0

3 years ago

Read 2 more answers

Answer all 3 parts of this essay question: A) (5pts) If you have a Periodic Table that is NOT color coded, describe where to loo

cestrela7 [59]

Answer:

A) If you look at the vertical rows you can tell how many electrons there are on the outer shell eg. Group 7 (eg. nitrogen) has 7 electrons on its outer shell

B) They are in the same group

C) Helium,Argon and Organessom are all Nobel gasses

8 0

4 years ago

Define principal focus of concave mirror.Where should we place a candle in front of a concave mirror to get an enlarged,erect im

Mars2501 [29]

Answer

The image formed by a concave mirror will be virtual, erect and magnified when the object is placed only between the pole and the principal focus of the mirror.

Hope it helped u if yes mark me brainliest!

Tysm!

3 0

4 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

4 years ago

A 16.9 kg monkey is swinging on a 5.32 m long vine. It starts at rest, with the vine at a 43.0° angle. How fast is the monkey mo

NemiM [27]

Answer:

v = 5.7554 m/s

Explanation:

First of all we need to know if the angle of the vine is measured in the horizontal or vertical.

To do this easier, let's assume the angle is measured with the horizontal. In this case, the innitial height of the monkey will be:

h₀ = h sinα

h₀ = 5.32 sin43° = 3.6282 m

As the monkey is dropping from the innitial point which is the suspension point, is also dropping from 5.32. Then the actual height of the monkey will be:

Δh = 5.32 - 3.63 = 1.69 m

In order to calculate the speed of the monkey we need to understand that the monkey has a potential energy. This energy, because of the gravity, is converted in kinetic energy, and the value will be the same. Therefore we can say that:

Ep = Ek

From here, we can calculate the speed of the monkey.

Ep = mgΔH

Ek = 1/2 mv²

The potential energy is:

Ep = 16.9 * 9.8 * 1.69 = 279.9

Now with the kinetic energy:

1/2 * (16.9) * v² = 279.9

v² = (279.9) * 2 / 16.9

v² = 33.12

v = √33.12

Hope this helps

3 0

3 years ago