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1 year ago

a car collides with a wall. compare the forces exerted by the car on the wall and the wall on the car

Physics

2 answers:

NeTakaya1 year ago

5 0

Mass multiplied by acceleration produces force.

The acceleration is (v - 0)/t in this situation, where t seems to be the time it takes automobile A to come to a stop. According to Newton's third law of motion, the automobile produces this turning force of the wall, however the wall, which really is static and indestructible, forces an equal force back on the car.

According to Newton's third law, each action has an equal and opposite response. On this basis, you may deduce that a car driving into a wall would exert force on the wall. However, since the wall did not move, the automobile receives an equivalent force, causing it to collapse.

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Helga [31]1 year ago

4 0

Answer:

Newton’s third law states that for every action there is an equal and opposite reaction. Based on this you can say that the car crashing into the wall would put force into the wall. However, because the wall didn’t move it gives an equal force back to the car, probably causing it to crumple.

Explanation:

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A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$

This force work as centripetal force

So $F=\frac{mv^2}{r}$

$4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}$

v = $=0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec$

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2 years ago

Salmon often jump waterfalls to reach their

PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

height of the waterfall, h = 0.432 m
distance of the Salmon from the waterfall, s = 3.17 m
angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

$h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s$

The minimum velocity of the Salmon jumping at the given angle is calculated as;

$X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s$

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

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1 year ago

A planet has a period of revolution about the Sun equal to and a mean distance from the Sun equal to R .T^2 varies directly as _

shutvik [7]

T² caries directly as R³ .

This is Kepler's 3rd law of planetary motion .

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2 years ago

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

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2 years ago

How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v

777dan777 [17]

The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
$W= qV_i - qV_f$
where q is the charge of the proton, $q=1 e = 1.6\cdot 10^{-19}C$ , with $e$ being the elementary charge, and $V_i = +125 V$ and $V_f = -55 V$ are the initial and final voltage.

Substituting, we get (in electronvolts):
$W=e(125 V-(-55 V))=180 eV$
and in Joule:
$W=(1.6 \cdot 10^{-19})(125 V-(-55V))=2.88 \cdot 10^{-17}J$

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3 years ago

a car collides with a wall. compare the forces exerted by the car on the wall and the wall on the car​

a car collides with a wall. compare the forces exerted by the car on the wall and the wall on the car