All categories

4 years ago

A tennis ball is dropped from 1.43 m above the

Physics

1 answer:

Rudiy274 years ago

4 0

Answer:

-5.29 m/s

Explanation:

Given:

y₀ = 1.43 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.43 m)

v = -5.29 m/s

You might be interested in

Malaria is a disease caused by a protozoan. This protozoan lives in the salivary glands of certain mosquitoes without causing th

Ray Of Light [21]

I think they are exhibiting symbiosis.

7 0

3 years ago

Given: g = 9.8 m/s 2 . A small glass of water is placed on carousel inside a microwave oven, at a radius 6.5 cm from the center.

pashok25 [27]

Answer:

0.17 degrees

Explanation:

From the question we are given the following:

acceleration due to gravity = $9.8 \frac{m}{s^{2} }$

radius (r) = 6.5 cm = 0.065 m

time (s) = 9.3 s

angle = ?

we first have to calculate the distance the water moves, and this is the circumference of the microwave

circumference = 2 x π x r = 2 x π x 0.065 = 0.408 m

Now we find the speed with which it moves at

speed = $\frac{distance}{time}$

= $\frac{o.408}{9.3}$ = 0.044

The next step is to find the centripetal acceleration

a = $\frac{v^{2} }{r}$

= $\frac{0.044^{2} }{0.065}$ = 0.03

being a vector quantity, acceleration has direction, the centripetal acceleration is the x component while the y component would be the acceleration due to gravity (g)

A = 0.03 x + 9.8 y

to find the angle we can apply tan θ = $\frac{ 0.03 }{9.8$

θ = 0.17 degrees

7 0

3 years ago

The type of force that will not cause a change in the speed of an object.

In-s [12.5K]

Explanation:

Balanced forces will cause no change in the speed of an object.

3 0

3 years ago

Roots that grow horizontally are called

Romashka [77]

I believe they are called Rhizomes.

8 0

4 years ago

Read 2 more answers

What is its diameter when the temperature is raised to 100 degrees Celsius? (b) What temperature change is required to increase

alexandr1967 [171]

The question is incomplete, the complete question is;

A spherical steel ball bearing has a diameter of 2.540cm at 25.00∘C.

(a) What is its diameter when it's temperature is raised to

100∘C?

(b) What temperature change is required to increase its volume by

1.000% ?

Answer:

a) 2.542 cm

b) 303.03°C

Explanation:

Given;

Diameter of the ball= 2.540cm

Initial temperature= 25.0°C

Final temperature= 100.0°C

Percentage increase in volume = 1.000%

Temperature coefficient of expansion for steel =11.0×10^−6/∘C

d2= d1[1 + α(T2-T1)]

d2= 2.540[1 + 11.0×10^−6(100-25)]

d2= 2.540[1 + 8.25×10^-4]

d2= 2.542 cm

From;

%V ×1/100 = V ×3α ×∆T/ V

Substituting values;

1.000 ×1/100= 3× 11.0×10^−6 × ∆T

∆T= 0.01/3× 11.0×10^−6

∆T= 303.03°C

4 0

3 years ago