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3 years ago

What is the momentum of a 950 kg car moving at 10.0 m/s?

Physics

2 answers:

KonstantinChe [14]3 years ago

6 0

What's 950 x 10 bro?

9500 kg m/s

OverLord2011 [107]3 years ago

6 0

Answer:

Momentum of the car, p = 9500 kg-m/s.

Explanation:

It is given that,

Mass of the car, m = 950 kg

Velocity of the car, v = 10 m/s

We need to find the momentum of the car. It is given by the product of mass and velocity. It is given by :

$p=m\times v$

$p=950\ kg\times 10\ m/s$

p = 9500 kg-m/s

So, the momentum of the car is 9500 kg-m/s. Hence, this is the required solution.

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Who is known as the "father of modern chemistry" because he first organized all known elements into four different

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Answer:

D. Henry Moseley

Explanation:

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3 years ago

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Karla Ayala pulls a sled on an icy road (dangerous!). Because of Karla's pull, the tension force is 151 N, and the rope makes a

skelet666 [1.2K]

Answer:

W = 1418.9 J = 1.418 KJ

Explanation:

In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:

W = F.d

W = Fd Cosθ

where,

W = Work Done = ?

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d = distance covered = 10 m

θ = Angle with horizontal = 20°

Therefore,

W = (151 N)(10 m) Cos 20°

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3 years ago

The smallest division value of electronic balance

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Answer:

0.1g to 0.0000001g hope it helps uu

4 0

2 years ago

The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,

madam [21]

Answer:

(b) 1:1

Explanation:

1. Formulae:

E = hf

E = h.v/λ

λ = h/(mv)

E = (1/2)mv²

Where:

E = kinetic energy of the particle
λ = de-Broglie wavelength
m = mass of the particle
v = speed of the particle
h = Planck constant

2. Reasoning

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

$\dfrac{m_\alpha}{m_p}=4$

For the kinetic energies you find:

$\dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}$

$\dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4$

Thus:

$\dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}$

$m_\alpha v_\alpha=m_pv_p$

From de-Broglie equation, λ = h/(mv)

$\dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1$

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3 years ago

If a lamp has a resistance of 50 ohms and is operated by a p.d. of 10V, find the current flowing through it

NemiM [27]

V=IR therefore I=V/R=10/50=0.2A therefore the current is 0.2 A

4 0

2 years ago