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DaniilM [7]
1 year ago
14

A net force is applied on a 100 kg rocket which causes the rocket to acceleration at 10 m/s2. The same net force is applied on a

25 kg rocket. What is the acceleration of the second rocket?
Physics
1 answer:
ELEN [110]1 year ago
8 0

Answer:

i dont know lo

Explanation:

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a partially inflated weather balloon has a volume of 1.56 * 10^3 L and a pressure of 98.9 lPa. What is the volume of the balloon
e-lub [12.9K]
To solve the problem, we enumerate all the given first. Then the required and lastly the solution.

Given:

V1= 1.56x10^3 L = 1560 L                P2 = 44.1 kPa
P1 = 98.9 kPa

Required:  V2

Solution:

Assuming the gas is ideal. Ideal gas follows Boyle's Law which states that at a given temperature the product of pressure and volume of a gas is constant. In equation,

PV = k

Applying to the problem, we have
 
P1*V1 = P2*V2
(98.9 kPa)*(1560 L) = (44.1 kPa)*V2
V2 = 3498.5 L

<em>ANSWER: V2 = 3498.5 L</em>
7 0
2 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
2 years ago
Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 4.00. What should be the length of B⃗ so that th
Soloha48 [4]

Answer:

B= \sqrt{65} ≅8.06

Explanation:

Using the Pythagorean theorem:

C^{2}= A^{2} + B^{2}

where C represents the length of the hypotenuse and A and B the lengths of the triangle's other two sides, we can find out the lenght of B assuming the value of the hypotenuse being 9 and A being 4.

9^{2}=4^{2} + B^{2}

81= 16+ B^{2}

81-16= B^{2}

B= \sqrt{65} ≅8.06

8 0
2 years ago
A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a
Maru [420]

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=mv_{2} +mv_{1}

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

4 0
2 years ago
If the mass of one of two particles is doubled and the distance between them is doubled, the force of attraction between the two
andre [41]
The force of attraction between the two particles will remain the same, because when mass is doubled, force of attraction is doubled. However, when distance between their centers is doubled, then force of  attraction is halved. As such double and half cancel out each other and force of attraction remains the same.
7 0
2 years ago
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