answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Julli [10]
2 years ago
11

HELP PLSSSSSSS ASAP FIRST CORRECT ANSWER WILL GET BRANLIST. A bunny accidentally knocks a basket of eggs off of a table. Luckily

the basket lands safely on the floor with all the eggs unbroken. Air resistance is negligible. What is the acceleration of the egg-basket system in midair? A)The acceleration cannot be determined without knowing how hard the basket was pushed. B) Acceleration is downward and less than g because the eggs are light. C) The acceleration fluctuates because of the rolling of the eggs. D) Acceleration is downward with a magnitude of g because the system is a projectile.
Physics
2 answers:
mezya [45]2 years ago
8 0

Answer:

A

Explanation:

hope this helps

Alenkasestr [34]2 years ago
7 0
The answer is A because it doesn’t not tell us how hard the egg was pushed. If it was pushed softly then it would probably not break. If it was pushed hardly then it would have probably broke.
You might be interested in
4. In a closed system consisting of a cannon and a cannonball, the kinetic energy of a cannon is 72,000 J. If the cannonball is
FromTheMoon [43]

Answer:

D an B

Explanation:

3 0
2 years ago
Read 2 more answers
A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

4 0
2 years ago
A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a
Yanka [14]

Answer:

Diameter of the cylinder will be d=2.998\times 10^4m

Explanation:

We have given young's modulus of steel E=207GPa=207\times 10^9Pa  

Change in length \Delta l=0.38mm

Length of rod l=500mm

Load F = 11100 KN

Strain is given by strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}

We know that young's modulus E=\frac{stress}{strain}

So 207\times 10^9=\frac{stress}{7.6\times 10^{-4}}

stress=1573.2\times 10^{-5}N/m^2

We know that stress =\frac{force}{artea }

So 1573.2\times 10^{-5}=\frac{11100\times 1000}{area}

area=7.055\times 10^{8}m^2

So \frac{\pi }{4}d^2=7.055\times 10^{8}

d=2.998\times 10^4m          

6 0
2 years ago
The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle
serg [7]

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we know that

m_1 = 50 g = 0.050 kg

m_2 = 3 kg

v_{1i} = v_{2i} = 0

v_{1f} = 400 m/s

now we will have

0.050(0) + 3(0) = 0.050(400) + 3(v)

20 + 3v = 0

v = - \frac{20}{3} = - 6.67 m/s

so gun will recoil with speed 6.67 m/s

6 0
2 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i
Debora [2.8K]

Answer:

(A) Total energy will be equal to 0.044\times 10^{-5}J

(b) Energy density will be equal to 0.0175J/m^3

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate A=\pi r^2=3.14\times 0.02^2=0.001256m^2

Distance between the plates d = 0.50 mm = 0.50\times 10^{-3}m

Permitivity of free space \epsilon _0=8.85\times 10^{-12}F/m

Potential difference V =200 volt

Capacitance between the plate is equal to C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F

(a) Total energy stored in the capacitor is equal to

E=\frac{1}{2}CV^2

E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J

(b) Volume will be equal to V=Ad, here A is area and d is distance between plates

V=0.001256\times 0.02=2.512\times 10^{-5}m^3

So energy density =\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3

7 0
2 years ago
Other questions:
  • Assuming the same current is running through two separate coils, why is it easier to thrust a magnet into a wire coil with one l
    6·2 answers
  • Jack pulls a sled across a level field by exerting a force of 110 n at an angle of 30 with the ground. what are the parallel and
    7·2 answers
  • You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af
    8·2 answers
  • Sarah invested 12000 in a unit trust 5 years ago, the value of the unit trust has increased by 7% per annum for the last 3 years
    10·2 answers
  • A sleepy student drops a calculator out of a window that's 20.7\text{ m}20.7 m20, point, 7, start text, space, m, end text off t
    10·1 answer
  • The gas tank of Dave’s car has a capacity of 12 gallons. The tank was 38 full before Dave filled it to capacity. It cost him $2.
    8·2 answers
  • Force F1 acts on a particle and does work W1. Force F2 acts simultaneously on the particle and does work W2. The speed of the pa
    9·1 answer
  • Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
    5·1 answer
  • A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, c
    6·2 answers
  • 11. Scientists put a sample of water into a sealed tank. Water can be a solid, liquid, or gas. At first, the water was a liquid.
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!