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IgorLugansk [536]
3 years ago
7

A baseball is released at rest from the top of the Washington Monument. It hits the ground after falling for 6 s. What was the h

eight from which the ball was dropped
Physics
1 answer:
alukav5142 [94]3 years ago
6 0

Answer:

Total height (s) = 176.4 m

Explanation:

Given:

Initial velocity (u) = 0 m/s

Time taken (t) = 6 sec

Acceleration due to gravity = 9.8 m/s²

Find:

Total height (s)

Computation:

s = ut + [1/2]gt²

s = (0)(6) + [1/2][9.8][6²]

s = 176.4 m

Total height (s) = 176.4 m

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What's the difference mass and weight
yanalaym [24]

Answer:

Mass doesn't change.

Weight is measured based on gravitational pull.

Explanation:

4 0
3 years ago
A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height
stira [4]

Answer:

9.5 kg m^2/s

Explanation:

The angular momentum of an object is given by:

L=mvr

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

L=(0.35)(9.0)(3.0)=9.5 kg m^2/s

4 0
3 years ago
An airplane traveling at 201 m/s makes a turn. What is the smallest radius of the circular path (in km) that the pilot can make
egoroff_w [7]

Answer:

the smallest radius of the circular path is 8.1 km

Explanation:

The computation of the smallest radius of the circular path is given below:

Given that

V = Velocity = 201 m/s

a_c = acceleration = 5 m/s^2

radius = ?

As we know that

a_c = V^2 ÷ r

5 = 201^2 ÷ r

r = 201^2 ÷ 5

= 8,080.2 g

= 8.1 km

Hence, the smallest radius of the circular path is 8.1 km

4 0
3 years ago
A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with
ivanzaharov [21]

Answer:

11.72 mm

Explanation:

The gravitational potential energy equals the potential energy of the spring hence

PE_{gravitational}=PE_{spring}

mgh=0.5kx^{2} where m is the mass of object, g is the acceleration due to gravity, h is the height, k is the spring constant and x is the extension of the spring

mgdsin\theta=0.5kx^{2} where \theta is the angle of inclination and d is the sliding distance

Making x the subject then

x=\sqrt {\frac {2mgdsin\theta}{k}}

Substituting the given values then

x=\sqrt{\frac {2\times 11\times 9.81\times 3\times sin 38}{2.9\times 10^{4}}}= 0.117240716\approx 11.72 mm

8 0
3 years ago
I need help with number 7 and 8. Gravity is 10N NOT 9.8N for these problems
noname [10]
7A 8A maybe I’m not sure sorry if it’s wrong.
8 0
3 years ago
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