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4 years ago

Please help with these i dont know how to do them

Physics

1 answer:

Svetradugi [14.3K]4 years ago

3 0

20.0M/S. IS THE HORIZONTAL VELOCITY OF THE BALL JUST BEFORE IT REACHES THE GROUND

12.2 SECONDS IS THE APPROXIMATE TOTAL TIME REQUIRED FOR THE BALL

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A thin rod rotates at a constant angular speed. Consider the tangential speed of each point on the rod for the case when the axi

ASHA 777 [7]

Answer:

v = R w

With this expression we see that for each point at different radius the tangential velocity is different

Explanation:

They indicate that the angular velocity is constant, that is

w = dθ / dt

Where θ is the radius swept angle and t the time taken.

The tangential velocity is linear or

v = dx / dt

Where x is the distance traveled in time (t)

In the definition of radians

θ = s / R

Where s is the arc traveled and R the radius vector from the pivot point, if the angle is small the arc (s) and the length (x) are almost equal

θ = x / R

We substitute in the speed equation

v = d (θ R) / dt

The radius is a constant for each point

v = R dθ / dt

v = R w

With this expression we see that for each point at different radius the tangential velocity is different

6 0

3 years ago

The magnitude of the gravitational acceleration at the surface of a planet of mass M and radius R is g. What is the magnitude of

maksim [4K]

2M/4R2=.5M/R2= .0.5 g

4 0

3 years ago

If a projectile is fired straight up at a speed of 10 m/s, the time it takes to reach the top of its path is about

UNO [17]

Answer:

A. The time it takes the projectile to reach the top of its path is about 1 second.

Explanation:

Hi there!

The equation of the velocity of a projectile fired straight up is the following:

v = v0 + g · t

Where:

v = velocity of the projectile.

v0 = initial velocity.

g = acceleration due to gravity (≅ -9.8 m/s² considering the upward direction as positive)

t = time.

When the projectile reaches the top of its path, its velocity is zero, then, using the equation of velocity, we can solve it for the time:

v = v0 + g · t

0 = 10 m/s - 9.8 m/s² · t

t = -10 m/s / -9.8 m/s²

t = 1.0 s

The time it takes the projectile to reach the top of its path is about 1 second.

7 0

3 years ago

If a model ship is scaled to the ratio of 1:1,200, what size is the real ship

Ipatiy [6.2K]

The answer I had got was C x

5 0

3 years ago

The radius of an atom of krypton (kr) is about 1.9 å. (a) express this distance in nanometers (nm). nm express this distance in

I am Lyosha [343]

Note:
1 A (armstrong) = 10⁻¹⁰ m
1 nm (nanometer) = 10⁻⁹ m

Given:
Radius of a krypton atom = 1.9 A = 1.9 x 10⁻¹⁰ m

Part (a)
$1.9\,A = (1.9\,A)*(10^{-10}\, \frac{m}{A})*( \frac{1}{10^{-8}} \frac{nm}{m}) } =0.019\,nm$
Answer: 0.019 nm

Part (b)
The diameter of a krypton atom = 2*1.9A = 3.8 A = 3.8 x 10⁻¹⁰ m.
The number of krypton atoms within a length of 1.0 mm is
$\frac{1.0\, mm}{3.8 \time 10^{-10}\, m} = \frac{10^{-3}\, m}{3.8 \times 10^{-10} \,m} =2.632 \times 10^{6}$

Answer: About 2.632 x 10⁶ atoms

Part (c)
The radius of a krypton atom is
1.9 A = (1.9 x 10⁻¹⁰ m)*(10² cm/m) = 1.9 x 10⁻⁸ cm
The volume of a krypton atm is
$\frac{4 \pi }{3} (1.9 \times 10^{-8} \, cm)^{3} = 2.873 \times 10^{-23} \, cm^{3}$

Answer: 2.873 x 10⁻²³ cm³

8 0

3 years ago