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AfilCa [17]
3 years ago
11

PLEASE HELP! FOR BRAINLIEST!

Physics
1 answer:
Rasek [7]3 years ago
8 0

Answer:

B is the answer

Explanation:

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Uma massa de 500 Kg desloca-se com velocidade 58 km por hora. Calcule o módulo de sua quantidade por movimento
Simora [160]

The momentum of the object is 8050 kg m/s

Explanation:

The momentum of an object is defined as

p=mv

where

p is the momentum

m is the mass

v is the velocity of the object

For the object in this problem, we have

m = 500 kg is its mass

v = 58 km/h is its velocity

Converting the velocity into m/s,

v=58 \frac{km}{h}\cdot \frac{1000 m/km}{3600 s/h}=16.1 m/s

Therefore now we can find the momentum of the object:

p=(500)(16.1)=8050 kg m/s

Learn more about momentum:

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7 0
3 years ago
An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k
alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

7 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
2 years ago
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Which situation best describes the act of reducing?
yawa3891 [41]
A - i think
paying bills online?
5 0
3 years ago
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The air in a kitchen has a mass of 60.0kg and a specific heat of 1505J/(kg°C).
BARSIC [14]
Your answer is 632,100J which is Choice D
8 0
3 years ago
Read 2 more answers
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