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Ymorist [56]
3 years ago
10

HELP PLEASE Fill in the blank: The ? of water is 1 g/cm3.

Physics
1 answer:
ololo11 [35]3 years ago
3 0
I think it’s density
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A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
What level of nitrates is considered clean water?
lana66690 [7]
The U.S. Environmental Protection Agency (EPA) standard for nitrate in drinking water is 10 milligrams of nitrate (measured as nitrogen) per liter of drinking water (mg/L). * Drinking water with levels of nitrate at or below 10 mg/L is considered safe for everyone.
6 0
3 years ago
Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims
777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force F = 12.3 N

Displacement d = 10.2 m

Mass of rocket m = 0.663 Kg

From work energy theorem,

  W = \Delta K

 W_{t} - Wd - W_{g} = KE

Where W_{t} = thrust work W_{g} = gravitational work

KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2

KE = 59.2 -Wd

After cutoff kinetic energy is converted into potential energy,

KE = Wd' + mg\Delta h

Put value of KE

59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5

Work done by drag force is given by,

Wd'+Wd =  59.2 -29.23

                 = 29.96 J

Therefore, the work done by the drag force is given by 29.96 J

5 0
3 years ago
If a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower?
solmaris [256]

Answer:

60.025m.

Explanation:

S= ut + at^2/2 (2nd equation of motion)

S= 0 + (9.8)(3.5)^2 /2 (free fall case, initial velocity = 0)

S = 4.9 x 12.25

S= 60.025 m.

Disclaimer: did math in my head, so you better double check with a calculator.

6 0
2 years ago
In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:
fenix001 [56]

Acceleration is given by:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) -20.5 m/s^2

Let's convert the quantities into SI units first:

u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s

v=1600 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =444.4 m/s

t = 4.0 min = 240 s

So the acceleration is

a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2

B) -3.8 m/s^2

As before, let's convert the quantities into SI units first:

u = 444.4 m/s

v=321 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =89.2 m/s

t = 94 s

So the acceleration is

a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2

C) -53.0 m/s^2

For this part we have to use a different formula:

v^2 - u^2 = 2ad

where we have

v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

a is the acceleration

d = 75 m is the distance covered

Solving for a, we find

a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2

3 0
3 years ago
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