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notsponge [240]
3 years ago
11

You toss a tennis ball straight upward. At the moment it leaves your hand it is at a height of 1.5 m above the ground, and it is

moving at a speed of 7.0 m/s.
(a) How much time does it take for the tennis ball to reach its maximum height? (s)

(b) What is the maximum height above the ground that the tennis ball reaches? (m)

(c) When the tennis ball is at a height of 2.4 m above the ground, what is its speed? (m/s)
Physics
1 answer:
ollegr [7]3 years ago
4 0

My answer was incorrect, please disregard.

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What is the density of a 15 g box measuring<br> 10 cm hy 5 cm by 2 cm?
Dvinal [7]

Answer:

The density of the box is 0.15 g/cm³

Explanation:

Given;

mass of the box, m = 15 g

dimension of the box, = 10 cm by 5 cm by 2 cm

Volume of the box, v = 10 cm x 5 cm x 2 cm = 100 cm³

The density of the box is given by;

Density = mass / volume

Density = 15 g / 100 cm³

Density = 0.15 g/cm³

Therefore, the density of the box is 0.15 g/cm³

8 0
3 years ago
A skydiver has reached terminal velocity—she now falls at a constant speed, so her acceleration is zero. Is there a net force on
Taya2010 [7]

Answer:

No, the net force on the skydiver is zero

Explanation:

According to Newton's Second Law, the net force on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass of the object

a is the acceleration

In this problem, the acceleration of the skydiver is zero:

a = 0

This implies that also the net force on the skydiver is zero, according to the previous equation:

F = 0

So, the net force on the skydiver is zero. This occurs because the air resistance, which points upward, exactly balances the force of gravity on the skydiver, acting downwards.

6 0
4 years ago
Is Salad dressing a compound ??
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7 0
3 years ago
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You have a grindstone (a disk) that is 95.2 kg, has a 0.399 m radius, and is turning at 93 rpm, and you press a steel axe agains
olya-2409 [2.1K]

Answer:

angular acceleration is -0.2063  rad/s²

Explanation:

Given data

mass m = 95.2 kg

radius r = 0.399 m

turning ω = 93 rpm

radial force N  = 19.6 N

kinetic coefficient of friction  μ = 0.2

to find out

angular acceleration

solution

we know frictional force that is = radial force × kinetic coefficient of friction

frictional force = 19.6 × 0.2

frictional force = 3.92 N

and

we know moment of inertia  that is

γ =  I ×α = frictional force × r

so

γ  = 1/2 mr²α

α  = -2f /mr

α  = -2(3.92) /95.2 (0.399)

α  = - 7.84 / 37.9848 = -0.2063

so angular acceleration is -0.2063  rad/s²

3 0
3 years ago
PLEASE HELP ASAP please like this is supper important
MaRussiya [10]
T(°C) = (T(°F) - 32) × 5/9

T(°F) = T(°C) × 9/5 + 32
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3 years ago
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