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2 years ago

HELP ME PLEASE URGENT WILL MARK AND 5 STARS

Physics

1 answer:

Artist 52 [7]2 years ago

7 0

Answer:

option (B) is the correct option.

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Does it take more force to slow something down than to speed it up.

ANTONII [103]

To bring something to a stop the same force that was applied to speed it up can be used to stop it. If a greater force is used it will stop quicker.

7 0

2 years ago

A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force

salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

$KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J$

Energy lost due to friction

$W=Fd=0.031\times 0.158=0.0048J$

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

$\frac{1}{2}mv^2=0.0052$

$\frac{1}{2}\times 0.00524\times v^2=0.0052$

v = 1.40 m/sec

7 0

2 years ago

What is the lupac name H3C – CHT the H₂ - CH - CH2 CH₂ CH₂ CH3

dimulka [17.4K]

Answer:??

Explanation:

5 0

2 years ago

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 31.0 nC , respecti

allsm [11]

Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q = −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

8 0

2 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

$r=\frac{mv}{qB}$ (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

$\omega=\frac{v}{r}$

Then, you solve the equation (1) in order to obtain v/r:

$\frac{v}{r}=\omega=\frac{qB}{m}$

Finally, you replace the values of the parameters:

$\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}$

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

$f=2\pi \omega=5.5*10^7Hz$

5 0

2 years ago