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3 years ago

An antibaryon composed of two antiup quarks

Physics

1 answer:

Bumek [7]3 years ago

6 0

An antibaryon composed of two antiup quarks and one antidown quark would have a charge of (2) −1e.

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N 1800kg car has an of 3.8m/s? What is it on the car? acceleration force acting

nevsk [136]

Answer:

6840 N

Explanation:

The force acting on the car can be found by using Newton's second law:

F = ma

where

F is the net force on the car

m is the mass of the car

a is its acceleration

For the car in this problem,

m = 1800 kg

$a=3.8 m/s^2$

Substituting,

$F=(1800)(3.8)=6840 N$

7 0

3 years ago

Two falling inflated balls of different masses land at the same time.

Mashcka [7]

Answer:

true two falling inflated balls of different mass lands at the smae time because gravity acts to both in a same way

8 0

3 years ago

A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87Â° from the direction of th

77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, $v=8.2\times 10^3\ m/s$

Angle between velocity and the magnetic field, $\theta=87$

Charge, $q=5.7\ \mu C=5.7\times 10^{-6}\ C$

Magnetic force, F = 0.002 N

The magnetic force is given by :

$F=qvB\ sin\theta$

B is the magnetic field

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}$

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

5 0

4 years ago

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th

katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

$F=2*10^{4}N$

Explanation:

From the question we are told that:

Mass $m=0.500kg$

Initial Velocity $V=15.0m/s$

Distance $x=2.80mm=>0.00280m$

Diameter $d=2.50mm=>0.00250m$

Length $l=6.00cm=>0.6m$

Generally the equation for Force is mathematically given by

$F=\frac{mv^2}{2d}$

$F=\frac{0.500*15^2}{2.80*10^{-3}}$

$F=2*10^{4}N$

6 0

3 years ago

The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,

Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

$Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N$

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

$a=Fr/m$

$a=-4291.875/1750\\a=-2.4525$

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

$v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m$

3 0

3 years ago