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Anna71 [15]
3 years ago
12

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp

eed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.
Physics
2 answers:
Alex787 [66]3 years ago
7 0

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

i want the answer i don't know

ipn [44]3 years ago
6 0

Answer:

Explanation:want answers

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The Geiger–Marsden experiment(s) (also called the Rutherford gold foil experiment) were a landmark series of experiments by which scientists discovered that every atom contains a nucleus where its positive charge and most of its mass are concentrated
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3 years ago
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A car moves with a speed of 72 km/h for 15 minutes and then with a speed of 80 km/h for the next 12 minutes. The total distance
sukhopar [10]

Answer:

The total distance covered by the car is 3,810.08 m

Explanation:

Given;

initial speed of the car, u = 72 km/hr = 20 m/s

initial time, t₁ = 15 minutes = 900 s

final speed of the car, v = 80 km/hr = 22.22 m/s

final time, t₂ = 12 minutes = 720 s

The acceleration of the car is given as;

a = \frac{v-u}{t_2 -t_1} \\\\a = \frac{22.22-20}{900-720}\\\\a = 0.0123 \ m/s^2

The total distance covered by the car is given as;

v² = u² + 2as

where;

s is the total distance covered by the car

22.22² = 20² + 2(0.0123)s

22.22²  - 20² = 2(0.0123)s

93.728 = 0.0246s

s = 93.728 / 0.0246

s = 3,810.08 m

Therefore, the total distance covered by the car is 3,810.08 m

8 0
3 years ago
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A car weighing 8000N is traveling at 45 m/s on a perfectly flat, frictionless road. If the driver slams on the brakes, how far w
laila [671]

Without friction, the car cannot stop...

3 0
2 years ago
A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio
Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be 8.21\times 10^3~N

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0
3 years ago
What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh
Goshia [24]

Answer:

a) R_s = 0.092

b) R_p = 0.085

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

R_s = 0.092

b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

R_p = 0.085

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3 years ago
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