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bezimeni [28]
3 years ago
9

Consider the following possibilities and select the correct choice.

Physics
1 answer:
dezoksy [38]3 years ago
4 0

Answer:

Tx not but mybe

Explanation:

for that reason its just trying to help

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ANSWER THIS FOR BRAINY CROWN B) ITS SO EASY!
Luda [366]

Answer:

300

Explanation:

3 0
3 years ago
Read 2 more answers
what is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5m/s
Dmitrij [34]

Hi there!

Recall Newton's Second Law:

\large\boxed{\Sigma F = ma}}

∑F = Net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are already given the mass and acceleration, so we can plug these values into the equation:

∑F = 1600 · 4.5 = 7200 N

7 0
3 years ago
(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a c
andreyandreev [35.5K]

Answer:

\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}

Explanation:

The formula for the surface of the circle is:

A(r) = \pi\cdot r^{2}

The rate of change of the spill area is obtained by deriving the previous formula in terms of time:

\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}

Finally, variables are replaced by known data:

\frac{dA}{dt} = 2\pi\cdot (39\,m)\cdot (1\,\frac{m}{s} )

\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}

4 0
4 years ago
Read 2 more answers
Got an F in Physical Science. HELP ME PLZZZ
Dmitriy789 [7]

i know can you plzz help me with this question im sorry i didt answer your question i just need hel.

7 0
3 years ago
A 4 kg block is launched up a 30° ramp with an initial speed of 5 m/s. The coefficient of kinetic friction between the block and
zlopas [31]

Answer:

The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

Explanation:

Given;

mass of block, m =  4 kg

coefficient of kinetic friction, μk = 0.25

angle of inclination, θ = 30°

initial speed of the block, u = 5 m/s

From Newton's second law of motion;

F = ma

a = F/m

Net horizontal force;

∑F = mgsinθ + μkmgcosθ

a = \frac{F_{NET}}{m} = \frac{mgsin \theta + \mu_kmgcos \theta}{m} \\\\a = gsin \theta + \mu_kgcos \theta\\\\a = 9.8sin (30) + 0.25*9.8cos(30)\\\\a = 4.9 + 2.1217\\\\a = 7.022 \ m/s^2

At the  top of the ramp, energy is conserved;

Kinetic energy = potential energy

¹/₂mv² = mgh

¹/₂ v² = gh

¹/₂ x 5² = 9.8h

12.5 = 9.8h

h = 12.5/9.8

h = 1.28 m

Height of the ramp is 1.28 m

Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;

v² = u² + 2ah

v²  = 5² + 2 x 7.022 x 1.28

v²  = 25 + 17.976

v²  = 42.976

v = √42.976

v = 6.56 m/s

Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

4 0
4 years ago
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