Help will mark brain list

A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.

Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

$a=\frac{v^{2} }{r}$ , where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

$a=\frac{3.5^{2} }{8.5} \\a=1.4$

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

7 0

3 years ago

Find the density of seawater at a depth where the pressure is 680 atm if the density at the surface is 1030 kg/m3. Seawater has

Pie

Answer:

$1060.41kg/m^3$

Explanation:

Bulk modulus is defined as the relative change in the volume of a body produced by a unit of compressive acting uniformly over its surface:

$B=\rho _o \frac{\bigtriangleup P }{\bigtriangleup \rho}$

Hence the density of the seawater at a depth of 680atm is calculated as:-

$\rho=\rho_o +\bigtriangleup \rho=\rho_o(1+\frac{\bigtriangleup P}{B})\\\\=1030 \times (1+ \frac{(680-1)\times10^5}{2.3\times 10^9})\\=1060.41kg/m^3$

4 0

3 years ago

Thandy is looking at two cells under the microscope.One is a human cheek cell and the other is a leaf mesophyll cell from a plan

Vanyuwa [196]

Cell are small and us human can't see them with our own eyes. it is in possible to see cell without a microscope

7 0

3 years ago

A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi

Marianna [84]

Answer:

Time period of horizontal Oscillation = T = 2 $\pi$ $\sqrt{\frac{m}{k} }$

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation = T = 2 $\pi$ $\sqrt{\frac{m}{k} }$

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

8 0

3 years ago

A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0

3 years ago