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2 years ago

Peter left Town A at 13:30 and travelled towards Town B at an

Physics

1 answer:

pochemuha2 years ago

4 0

Answer:

Explanation: From 13:30 to 15:00, it past: 1 h 30 mins = 1.5

Then, the distance covered by Peter: 40x1.5= 60 miles

From 13:45 to 15:00, it pasts; 1 h 15min =1.25

Then, the distance covered by Philip. 30 x 1.25 = 37.5 miles

Lastly, the distance between them: 60-37.5= 22.5 miles

So the answer is 22.5

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A student states that 203.9 g of table sugar will form a saturated solution in 100 g of water.

riadik2000 [5.3K]

The temperature and the solubility of sugar at that temperature

Explanation:

The amount of substance which can be dissolved in the solvent depends on the temperature.

As the temperature increases, more substance can be dissolved.

A solution is saturated if any more of the solute cannot be dissolved in the solution at the given temperature

Hence we need to know the temperature and also the amount of substance which can be dissolved(solubility) at the same temperature

a) the statement given in option A is correct

b) molar mass has no correlation with the substance's solubility and hence option b is not correct

c) The percent by volume of the solution is not needed to find if the solution is saturated and hence option c is not correct

3 0

2 years ago

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0

2 years ago

Read 2 more answers

Most binary systems with an invisible companion contain a large, bright star and a small, dim star hidden by the light of its la

Vlada [557]

Answer:

Brain signals are converted

8 0

2 years ago

Батискаф витримує тиск 60 МПа. Чи можна провести дослідження

Elanso [62]

1) Yes

2) $6.34\cdot 10^9 N$

Explanation:

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

$p=p_0+\rho g h$

where

$p_0 = 101,300 Pa$ is the atmospheric pressure

$\rho$ is the fluid density

$g=10 m/s^2$ is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

$\rho=1030 kg/m^3$ is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

$p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa$

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

$p=\frac{F}{A}$

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

$p=56.1\cdot 10^6 Pa$ is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

$A=4\pi r^2$

Therefore, we can find the force exerted on it by re-arranging the previous equation:

$F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N$

6 0

2 years ago

What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?

DIA [1.3K]

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

$F=kx$

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

$k=\frac{15.0-0}{10.0-0}=1.5 N/m$

Therefore when the spring has a elongation of $x=3.0 m$ , its potential energy is

$E=\frac{1}{2}(1.5)(3.0)^2=6.8 J$

Learn more about potential energy:

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3 0

2 years ago