Which chemical substance has the highest melting point?

If an object is moving, can it be considered "at rest"?

Irina-Kira [14]

No..........................................

7 0

4 years ago

Need help asap pls

Zinaida [17]

B. third

for every action there is a reaction*

6 0

2 years ago

A small convex mirror is placed 60 cm from the pole and on the axis of a large concave mirror of radius of curvature 200 cm. The

umka2103 [35]

Answer:

Calculate 'R' of convex mirror and height of the real image

the radius of the convex mirror is 48cm

Explanation:

Distance between convex and concave mirror is =60cm

Radius of the concave mirror (R) = 200cm

For the concave mirror, u = ∞

V = {R}/{2}=100cm

The object for the convex mirror and the final image is on the pole of the concave mirror, and distance between convex and concave mirror is 60cm

u_1=60-100 =-40cm

Object will be behind the convex mirror

1/f=1/40+1/60

f=24cms

the radius of the convex mirror is 48cm

6 0

3 years ago

X-rays cannot pass through Earth's atmosphere. Which of these is the best location to place a telescope used to observe x-rays f

Amiraneli [1.4K]

Mountains, tops of buildings, and high-flying aircraft are all part of Earth's atmosphere, no matter how high they are. On the other hand, space doesn't belong to our atmosphere, it is outside of it. Having this in mind, the best location to place a telescope used to observe x-rays from stars is in space.

7 0

3 years ago

Read 2 more answers

a baseball is hit straight up into the air. If the initial velocity was 20 m/s, how high will the ball go?How long will it be un

AleksAgata [21]

1) The motion of the ball is an uniformly accelerated motion, with constant acceleration equal to $g=-10 m/s^2$ (the negative sign means it is directed towards the ground).

For an uniformly accelerated motion, we can use the following relationship:
$2gh=v_f^2-v_i^2$
where h is the maximum height reached by the ball, $v_i = 20 m/s$ is its initial velocity and $v_f$ the velocity of the ball when it reaches the maximum height. But $v_f=0$ (when the ball reaches the maximum height, it stops before going down, so its velocity at that moment is zero), so we can use the relationship to calculate h, the maximum height:
$h=- \frac{v_i^2}{2g} =- \frac{20 m/s)^2}{2 \cdot (-10 m/s^2)} =20 m$

2) We can find the time the ball takes to return to the ground by requiring that the space covered by the ball returns to zero:
$S(t)=0$
where for an uniformly accelerated motion,
$S(t)=v_i t + \frac{1}{2} gt^2 =0$
By solving this, we have two solutions: one is t=0, which corresponds to the moment the player hits the ball, the second one is
$t=- \frac{2 v_i}{g}=- \frac{2 \cdot 20 m/s}{-10 m/s^2}=4 s$
so, the ball returns to the ground after 4 s.

3) The velocity of the ball when it returns to the ground is given by:
$v(t) = v_i +gt=20 m/s + (-10 m/s^2)(4 s)=-20 m/s$
where the negative sign means the ball is going downwards.

8 0

3 years ago