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anzhelika [568]
3 years ago
5

PLEASE THINK ABOUT THIS AND EXPLAIN FULLY :)

Physics
1 answer:
Free_Kalibri [48]3 years ago
6 0

Explanation:

It is given that,

Mass of object, m = 2 kg

Initial velocity, u = 3 m/s (east)

Final velocity, v = - 7 m/s (west)

The Impulse can be calculated using the change in momentum of an object i.e.

J = m(v-u)

J=2\ kg(-7\ m/s-3\ m/s)

J = -20 kg-m/s

So, the Impulse of this object is 20 kg-m/s but the direction is opposite. Hence, statement (1) is correct i.e. Student #1: "The impulse is equal to the change in momentum, which is (2 kg)(3 m/s + 7 m/s) = 20 kg m/s."

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A 2000 kg car slams on the brakes and slows down at a rate of -10 m/s2. How much force are the brakes applying?
Gemiola [76]

Answer:

-20,000N

Explanation:

Force (N) = mass (kg) x acceleration (m/s²)

So,

Force = 2000 x -10

= -20,000N (Newtons)

5 0
2 years ago
How do I calculate the efficiency of motor
zepelin [54]

Answer:

the formula is efficiency = output / input × 100%

7 0
3 years ago
you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and
GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb  

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s  

46 mph*1.46667=67.4666668 ft/s

Momentum_B=1125*67.4666668 ft/s

Initial momentum of car A is given by

Momentum_A=1515v_a where v_a is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

1515v_a-(1125*67.4666668 ft/s)

The common velocity is represented as v_c hence after collision, the final momentum is

Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

1515v_a-(1125*67.4666668 ft/s)= 2640v_c

The acceleration of two cars a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}

From kinematic equation

v^{2}=u^{2}+2as hence

v^{2}-u^{2}=2as

0^{2}-(v_c)^{2}=2*-24.1275*19.5

v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s

Substituting the value of v_c in equation 1515v_a-(1125*67.4666668 ft/s)= 2640v_c

1515v_a-(1125*67.4666668 ft/s)= 2640*30.67

1515v_a=(1125*67.4666668 ft/s)+2640*30.67

v_a=\frac {156868.8}{1515}=103.5438 ft/s

\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph

3 0
3 years ago
Four children pull on the same stuffed toy at the same time yet there is no net force on the toy. How is this possible.
Stels [109]

Answer:

Net force is Zero.

Explanation:

If all forces that are equal and opposite are exerted on an object the resulting force will be Zero.

3 0
2 years ago
Read 2 more answers
What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa
dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}

E=3.315\times10^{-17}\ J

(b).  We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}

E=6.709\times10^{-21}\ J

(c). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}

E=3.315\times10^{-11}\ J

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}

E=6.709\times10^{-9}\ J

Hence, This is the required solution.

6 0
3 years ago
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