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Darya [45]
2 years ago
10

What is the estimated density of the golf ball? Record your answer to the nearest hundreth.

Physics
1 answer:
Hitman42 [59]2 years ago
7 0

The estimated density of the golf ball is  170  kg/m³

What is density?

Density is the ratio as Mass per unit Volume.

In displacement method,  

First , we measuring the volume of water displaced by an object which tell us the volume of the object then we will use the physical balance to determine its mass.

Then calculate the density by dividing the mass by the volume.

Here,  D = m/V

By displacement method , the estimated volume of golf ball is 100 cm³  and estimated mass is 600 g

Then ,

Density =  100 cm³ / 600 g = 0. 17 g/ cm³ = 170  kg/m³

So the estimated density of golf ball is  170  kg/m³

For more content related to Density visit here;

brainly.com/question/12006917

#SPJ1

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A lizard accelerates from 2 m/s to 10 m/s in 4 seconds. what is the lizard average acceleration
VLD [36.1K]

Acceleration = (change in speed) / (time for the change)

change in speed = (ending speed) - (starting speed)

change in speed = (10 m/s) - (2 m/s)  =  8 m/s

Acceleration = (8 m/s) / (4 sec)

Acceleration = (8/4) (m/s²)

<em>Acceleration = 2 m/s²</em>

8 0
3 years ago
a car travel the first 20km with a speed of 40km/h and the next 40km with a speed of 80km/h . find the average speed​
REY [17]

Answer:

average speed is 60km/h

Explanation:

you sum up the speed attained in each distance covered and divide it by 2 to get your answer

3 0
3 years ago
A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total
Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

v_2 = v - 1

now we know that total time of the motion is 19 hr

so we will have

t_1 = \frac{90}{v}

t_2 = \frac{90}{v - 1}

so we will have

t_1 + t_2 = 19 hr

\frac{90}{v} + \frac{90}{v-1} = 19

90(2v - 1) = 19(v^2 - v)

19 v^2 - 199 v + 90 = 0

by solving above equation we have

v = 10 mph

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

5 0
3 years ago
5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)
jolli1 [7]

Answer:

(B) 13.9 m

(C) 1.06 s

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

3 0
3 years ago
A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w
myrzilka [38]
The centripetal acceleration is given by
a_c =  \frac{v^2}{r}
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, a_c = 15.625 m/s^2 and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s
8 0
3 years ago
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