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Semenov [28]
3 years ago
13

9.

Physics
1 answer:
zepelin [54]3 years ago
8 0
A I believe it would be true
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Lexy throws a dart with an initial velocity of 25 m/s at an angle of 60° relative to the ground. what is the approximate vertica
Serjik [45]
An initial velocity is:
v o = 25 m/s
The vertical component of the initial velocity:
v o y = v o * sin 60° =
= v o * √3 / 2 = 25 m/s * √3 / 2 = 21.65 m/s
Answer:
The approximate vertical component of the initial velocity is 21.65 m/s.
7 0
3 years ago
Read 2 more answers
A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained
mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

 m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm

3 0
2 years ago
Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
seraphim [82]

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

7 0
3 years ago
1 point
Svetllana [295]

Answer:

Beta 17,000K, bc warmer is more blue

5 0
3 years ago
A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass
iragen [17]

Answer:

64.5 cm

Explanation:

30 * 80 + x * 110 = 50* 190 => x = 64.5

5 0
3 years ago
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