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Fudgin [204]
3 years ago
13

Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2.

Gun #1 weighs three times as much as gun #2. The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is:
a) 2:3
b) 3:2
c) 2:1
d) 3:1
e) 6:1
Physics
1 answer:
IrinaVladis [17]3 years ago
7 0

Answer:

option C

Explanation:

Let mass of the bullet be m and velocity be v

mass of gun be M and bullet be V

now,

using conservation of momentum for gun 1

(M+m) V' = 2 mv + 3 MV

V' = 0

3 M V = - 2 mv

momentum of gun 1 =- 2 mv---------(1)

now for gun 2

(M+m) V' = mv + MV

V' = 0

M V = - mv

momentum of gun 1 = -mv-----------(2)

dividing equation (1) by (2)

\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}

\dfrac{P_m1}{P_m2} = \dfrac{2}{1}

the correct answer is option C

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What is the objective of playing basagang palayok?the goal what is the goal of playing basang palayok ​
liubo4ka [24]

Answer:

Explanation:

At first glance, it looked like a group of kids-at-heart playing traditional Filipino games under the sun. But up close the games were played by public school teachers staging a unique protest to demand government to prioritize their welfare.

Some 100 members of the Teacher’ Dignity Coalition (TDC) from Metro Manila staged what they called “Palarong Pampista” at the Plaza Miranda in Manila yesterday to demand salary increases this year.

The protest is part of the “Protest de Mayo” launched by TDC earlier this month to call for a P10,000 across-the-board increase in salaries of government teachers and the scrapping of the performance-based bonus of the Aquino government which they described as “deceptive, unfair and divisive.”

During the protest, teachers played traditional fiesta games like Palo Sebo and Basagang Palayok “to demonstrate their sacrifices for the country despite the inadequate compensation from the government.”

TDC National Chairperson Benjo Basas said Palo Sebo and Basagang Palayok were symbols of how the government is treating public school teachers.

Palo Sebo, a popular fiesta game in which contestants climb a greased bamboo in order to get the money prize at the top, is the same as the government’s performance-based bonus or PBB, Basas said.

“It promises a cash prize, but teachers and employees need to fight and pull others down in order to be on top,” he explained.

The Basagang Palayok is also another popular fiesta game where the players’ objective is to hit the hanging pot with prizes. Basas said that in this game, “the player is blindfolded, and like the P10,000 salary increase demand, the prize is uncertain and the players must make all the effort and pass the obstacles.”

Earlier, Education Secretary Armin Luistro said that the Department of Education (DepEd) “recognizes the right of our teachers to raise their concerns publicly.”

“We will not object to any measure that will help public school teachers,” said Luistro. “As long as there is adequate funding, a raise in teachers’ salaries will be welcome,” he said.

However, while DepEd expressed support to the plight of public school teachers for salary increase, Luistro enjoined “certain groups not to take any action that greatly affects the delivery of basic services to our learners.”

TDC, a 30,000-strong group, has set several other unique mass actions until the resumption of classes in June 2 in all public elementary and secondary schools nationwide.

5 0
3 years ago
"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
MrRissso [65]

Answer:

life (N) of the specimen is 117000  cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f  = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = \frac{(f\times Su)^2}{Se}     ......................1  

put here value and we get

a = \frac{(0.82\times 120)^2}{60}  

a = 161.4  kpsi

so coefficient value (b) will be

b = -\frac{1}{3}log\frac{(f\times Su)}{Se}  

b =  -\frac{1}{3}log\frac{(0.82\times 120)}{60}  

b = −0.0716

so here number of cycle N will be  

N =  (\frac{ \sigma a}{a})^{1/b}

put here value  and we get

N =  (\frac{ 70}{161.4})^{1/-0.0716}

N = 117000

so life (N) of the specimen is 117000  cycles

7 0
3 years ago
11. Reflecting telescopes are popular because they're _______ than a refracting telescope.
Anon25 [30]
I do not know what the school expects as an answer, but advantage of reflecting telescopes is that there is only one major reflecting surface, so it is quite easy to create a 6 or 8 inch telescope by an amateur, after adding on a prism and an eyepiece.  (a microscope eyepiece could be used).
MY answer would be "easier to build".  (it still takes tens of hours to grind and polish the single plane surface to a parabolic surface).

Electromagnetic waves all have the same velocity in the same medium.  However, since frequencies vary widely, so do wavelengths.

4 0
3 years ago
Read 2 more answers
Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
A pulley with the radius of 10.0 cm was connected to a motor with a massless
kogti [31]

Answer:

(i) -556 rad/s²

(ii) 17900 revolutions

(iii) 11250 meters

(iv) -55.6 m/s²

(v) 18 seconds

Explanation:

(i) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

α = (10000 − 15000) / 9

α ≈ -556 rad/s²

(ii) Constant acceleration equation:

θ = θ₀ + ω₀ t + ½ αt²

θ = 0 + (15000) (9) + ½ (-556) (9)²

θ = 112500 radians

θ ≈ 17900 revolutions

(iii) Linear displacement equals radius times angular displacement:

s = rθ

s = (0.100 m) (112500 radians)

s = 11250 meters

(iv) Linear acceleration equals radius times angular acceleration:

a = rα

a = (0.100 m) (-556 rad/s²)

a = -55.6 m/s²

(v) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

-556 = (0 − 15000) / t

t = 27

t − 9 = 18 seconds

8 0
3 years ago
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