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Vera_Pavlovna [14]
4 years ago
6

#3: a container has the dimensions of 30 cm x 50 mm x 0.2 m. the density of its contents is 2.5 g/cm3. what is the mass of the s

ubstance in kilograms? (d = m/v)
Physics
2 answers:
Dafna11 [192]4 years ago
7 0

Answer: m = 3.8

Explanation:

WARRIOR [948]4 years ago
4 0
Good afternoon!

We calculate the volume of the container in cm³. To do that, we must put the units in cm:

30 cm → 30 cm
50 mm → 5 cm
0.2 m → 20 cm

The volume is:

V = 30 . 5 . 20

V = 3000 cm³

Now, we calculate the mas with the formula:

m = dV

m = 2.5 · 3000

m = 7500 g

Dividing by 1000, we have the mass in kg:


m = 7.5 kg
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6. A 73-kg woman stands on a scale in an elevator. The
olga nikolaevna [1]

Answer:

b

Explanation:

n = m(g +a)

n= normal force (N)

m=mass (kg)

g=acceleration of gravity

a= acceleration of elevator

rearrange:

a= n/m - g

a= (810 N/73 kg) - 9.8 m/s ^2

a= 1.3 m/s ^2 up

and the acceleration is upwards bc her weight is less than the scale reading

3 0
3 years ago
A spherical balloon is deflating at 10 cm3/s. At what rate is the radius changing when the volume is 1000π cm3 ? What is the rat
just olya [345]

Answer:

1.dr/dt=0.0096cm/s

2. dA/dt=2.19cm^2/s

Explanation:

A spherical balloon is deflating at 10 cm3/s. At what rate is the radius changing when the volume is 1000π cm3 ? What is the rate of change of surface area at this moment?

for this question, we need to analyze the parameters we know

V=volume of the spherical balloon 1000π cm3

volume of the sphere=\frac{4}{3} \pi r^{3}

1000π=4/3πr^3

dividing both sides by 4

250*3=r^3

r=9.08cm, the radius of the balloon

dv/dt=dv/dr*dr/dt...................................1

dv/dr  ,means

V=\frac{4}{3} \pi r^{3}

dv/dr=4*pi*r^2

dv/dt=10 cm3/s

from equ 1

10=4*pi*9.08^2*dr/dt

10=1036 dr/dt

dr/dt=10/1036

dr/dt=0.0096cm/s

2. to find the rate at which the area is changing we have,

dA/dt=dA/dr*dr/dt

area of a sphere is  4πr^2

differentiate a with respect to r, radius

dA/dr=8πr

dA/dt=8πr*0.0096

dA/dt=8*pi*9.08*0.0096

dA/dt=2.19cm^2/s

is the rate of change of the surface area

7 0
3 years ago
A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.
Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0
3 years ago
Please help on this one?
Setler79 [48]

I may be wrong but I think the answer is open. Hope this helps.

8 0
4 years ago
Why did scientists agree to use one system of measurement
Zolol [24]
To make it easier to share data and experimental results with other scientists from all over the world. 
7 0
4 years ago
Read 2 more answers
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