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3 years ago

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a car with a mass of 100 kg is stopped on the side of the road after getting a flat tire. the two people that were riding in the

car get out and begin to push the car from rest to a nearby gas station. The car travels 50 meters in 40 seconds. Determine the speed of the car in these 40 seconds.

1 answer:

notsponge [240]3 years ago

3 0

Answer:

Please mark me brainliest and thank me and rate me

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An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit

lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is $1.127\times 10^7m/s$

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

$m_1v_1=m_2v_2$

where,

$m_1\text{ and }v_1$ = Initial mass and velocity

$m_2\text{ and }v_2$ = Final mass and velocity

We are given:

$m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s$

Putting values in above equation, we get:

$238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s$

Hence, the velocity of released alpha particle is $1.127\times 10^7m/s$

4 0

3 years ago

If an object has a net positive charge of 4.0 coulombs, the object possesses (1) 6.3 × 10^18 more electrons than protons (2) 2.5

Readme [11.4K]

Answer:

(4) $2.5 \times 10^{19}$ more protons than electrons

Explanation:

Total positive change on the object is given as

$Q = 4 C$

now by the law of quantization of charge we know that the charge on any body is always in form of integral multiple of charge of an electron.

So we will have

$Q = N e$

here we know that

$4 = N(1.6 \times 10^{-19})$

$N = 2.5 \times 10^{19}$

Now we know that the object is positively charged so here electrons must be less in numbers than protons

so correct answer will be

(4) $2.5 \times 10^{19}$ more protons than electrons

3 0

4 years ago

Un cuerpo de 60 kg se encuentra a una distancia de 3.5 m del otro cuerpo, de manera que entre ellos se produce una fuerza de 6.5

aev [14]

Answer:

1989.6Kg

Explanation:

The computation of the mass of the other body is given below:

As we know that

F = G × m1 × m2 ÷ r²

Here the G would have the constant value i.e. 6.67 × 10^-11Nm² / kg².

Now

6.5 × 10^-7N = 6.67 × 10^-11Nm² / kg² × 60Kg × m2 / (3.5m) ²

m2 = (F × r²) / (G × m1)

m2 = (6.5 × 10^-7N × (3.5m) ²) ÷ (6.67 × 10^-11Nm² / kg² × 60Kg)

= 1989.6Kg

7 0

3 years ago

What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the

timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$

So, M.I. of disk B = $3I_0$

Angular velocity of A = $\omega_0$

So the kinetic energy of the disk A = $\frac{1}{2}I_0\omega^2$

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4}$

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J$

Therefore, the maximum initial K.E. = 3066.67 J

3 0

3 years ago

Why is the specific heat of water important to our planet?

katen-ka-za [31]

Because heat keeps us warm and water hydrates the planet i guess?????
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3 0

3 years ago

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