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velikii [3]
2 years ago
10

The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob

ject will be dropped with no initial speed. The objects are initially 179 cm above the ground and separated by 142 cm. What will be the difference in the landing locations of the two objects?
Physics
1 answer:
Crazy boy [7]2 years ago
4 0

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

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zlopas [31]
-40 c = -40 f but k would be 233.15
6 0
3 years ago
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
2 years ago
An airplane has a starting velocity of 300m/s. It then accelerates at a rate of 45m/s2 for a time of 10s. What is it's final vel
Olenka [21]
A = (v - u) / t

a = acceleration
v = final velocity
u = initial velocity
t = time

45 = (v - 300) / 10

45 × 10 = v - 300

450 + 300 = v

v = 750 m/s

Hope this helps!

P.S. Let me know if you need an explanation
8 0
2 years ago
An Austin volleyball player bumps a 5 kg ball into the air. It reaches a height of 2.8 meters. How fast was the ball going as it
KATRIN_1 [288]

Answer:

v = 7.4 m/s

Explanation:

Given that,

Mass if a volleyball, m = 5 kg

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We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 2.8} \\\\=7.4\ m/s

So, the required speed is 7.4 m/s. Hence, the correct option is (b).

6 0
3 years ago
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Very specific alignment of the Sun, Earth, and Moon. If the Moon is lined up precisely with the Sun from the Earth's point of view, the Moon will block Sunlight from reaching the Earth, causing a solar eclipse.
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