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Kazeer
3 months ago
15

A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil

e lands on a hill 7.5m away. The height at which the projectile lands is most nearly
A- 1.78m
B- 3.10m
C- 5.34m
D- 6.68m
E- 12.0m
Physics
2 answers:
ValentinkaMS [3.4K]3 months ago
7 0

Answer:gggffdss

Explanation:cc bbhfewaz

Softa [3K]3 months ago
6 0

vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7Answer:

Explanation:

The range is specified as

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Two experiments are performed on an object to determine how much the object resists a change in its state of motion while at res
inna [3103]

Answer:

The first experiment measures inertial mass, while the second experiment measures gravitational mass.

Explanation:

A student conducts two different experiments to observe resistance to changes in motion, both when at rest and in motion.

In the initial experiment, an object is forcefully pushed against a flat surface while its speed is tracked by a sensor. This setup involves work done against the object's inertia, identifying the mass as inertial mass.

Conversely, in the subsequent experiment, the object is lifted or thrown upward with an applied force and the speed is recorded. Here, the mass refers to gravitational mass, as the work performed combats gravity or the object's weight.

5 0
3 months ago
When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic
ValentinkaMS [3465]
Total energy associated with a spring:
E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2

When x = 0.5a:
\frac{1}{2} k \frac{a}{2} ^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2 \\ \frac{1}{2} mv^2 = \frac{1}{2} ka^2 - \frac{1}{8} ka^2 = \frac{3}{8} ka^2

The ratio:
\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} = \frac{3}{4}
5 0
2 months ago
Read 2 more answers
A proton (mass = 1.67 10–27 kg, charge = 1.60 10–19 C) moves from point A to point B under the influence of an electrostatic for
serg [3582]

Answer:

20.353125 V

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

q = Charge of proton = 1.6\times 10^{-19}\ C

v_A = Velocity of proton at point A = 50 km/s

v_B = Velocity of proton at point B = 80 km/s

The relationship derived from energy conservation is as follows:

\dfrac{1}{2}m(v_B^2-v_A^2)=q(V_B-V_A)\\\Rightarrow V_B-V_A=\dfrac{1}{2q}m(v_B^2-v_A^2)\\\Rightarrow V_B-V_A=\dfrac{1}{2\times 1.6\times 10^{-19}}\times 1.67\times 10^{-27}(80000^2-50000^2)\\\Rightarrow V_B-V_A=20.353125\ V

The determined potential difference is 20.353125 V

3 0
3 months ago
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