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meriva
3 years ago
8

A blank tells you what the different symbols and lines represent on a map

Physics
1 answer:
Oduvanchick [21]3 years ago
3 0

A legend on a map tells you what different symbols mean

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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 2.0 s later. You ma
malfutka [58]

Answer:

4.9 m/s

Hope this helps! c:

Explanation:

3 0
4 years ago
A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.
g100num [7]

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle=34^{\circ}

Range=240 m

We know that Range =\frac{u^2sin2\theta }{g}

240=\frac{u^2sin(68)}{9.81}

u=50.39 m/s

(b)maximum height of projectile is given by

H=\frac{u^2sin^2\theta }{2g}

H=\frac{50.39^2(sin34)^2}{2\times 9.81}

H=40.46 m

3 0
4 years ago
Read 2 more answers
You travel in a car 30 miles north to ava and home again. what is your distance? __________________________________________ what
Luden [163]
If you travel 30 miles somewhere and then come home again your distance is 60 miles.  Your displacement is 0 because it is the ending position minus the beginning position, which are the same place (home).  In other words, displacement is a vector and distance is a scalar.
4 0
3 years ago
A child sits on the edge of a spinning merry go round that has a radius of 1.5 . The child’s speed is a 2m/s. What is the child’
Leto [7]

Answer: 2.67 m/s^2

Explanation:

Centripetal acceleration is defined as v^2/r; in this case, it's 2^2/1.5, which is 2.67.

3 0
3 years ago
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike
Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=9.50\times 10^{3} kg

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

mu+ MU=mv+ MV

Substitute the values

9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V

3.61i=2.375j+0.150V

3.61 i-2.375j=0.150V

V=\frac{1}{0.150}(3.61 i-2.375j)

V=24.07i-15.83j

Magnitude of velocity of stone

=\sqrt{(24.07)^2+(-15.83)^2}

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

\theta=tan^{-1}(\frac{y}{x})

=tan^{-1}(\frac{-15.83}{24.07})

\theta=tan^{-1}(-0.657)

=33.3 degree below the horizontal

(b)

Initial kinetic energy

K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2

K_i=685.9 J

Final kinetic energy

K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2

=\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2

K_f=359.12 J

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0
3 years ago
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