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3 years ago

Please answer this question

Physics

1 answer:

nata0808 [166]3 years ago

5 0

Answer:

Its sure 2nd gare beaciuse

Explanation:

of tobi was obito was tobi his som came and start doing enginner but he faild so i gave him answer he repiled oh baby baby oh

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Basking in the sun, a 1.10 kg lizard lies on a flat rock tilted at an angle of 15.0° with respect to the horizontal. What is the

Ray Of Light [21]

10.4 N

Given

m = 1.10 kg

θ = 15.0°

g = 9.81 m/s2

Solution

Fnet, y = ΣF y = Fn − Fg, y = 0

Fn = Fg, y = Fg

cosθ = mgcosθ

Fn = (1.10 kg)(9.81 m/s2

)(cos15.0°) = 10.4 N

3 0

3 years ago

If the jet is moving at a speed of 1140 km/h at the lowest point of the loop, determine the minimum radius of the circle so that

AnnZ [28]

To solve this problem we will apply the concepts related to the centripetal Force and the Force given by weight and formulated in Newton's second law. Through the two expressions we can find the radius of curve made in the hand. To calculate the normal force, we will include the concepts of sum of forces to obtain the net force on the body at the top and bottom of the maneuver. The expression for centripetal force acting on the jet is

$F_c = \frac{mv^2}{r}$

According to Newton's second law, the net force acting on the jet is

F = ma

Here,

m = mass

a = acceleration

v = Velocity

r = Radius

PART A ) Equating the above two expression the equation for radius is

$\frac{mv^2}{r} = ma$

$r = \frac{v^2}{a}$

Replacing with our values we have that

$r = \frac{(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{7(9.8m/s^2)}$

$r = 1.462*10^3m$

PART B )

- The expression for effective weight of the pilot at the bottom of the circle is

$N = mg +\frac{mv^2}{r}$

$N = (69kg)(9.8m/s^2)+\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}$

$N = 5408.87N$

Note that the normal reaction N is directed upwards and gravitational force mg is directed downwards. At the bottom of the circle, the centripetal force is directed upwards. So the centripetal force is obtained from the gravitational force and the normal reaction.

- The expression for effective weight of the pilot at the top of the circle is

$N = mg -\frac{mv^2}{r}$

$N = (69kg)(9.8m/s^2)-\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}$

$N = 4056.47N$

Note that at the top of the circle the centripetal force is directed downwards. So the centripetal force is obtained from normal reaction and the gravitational force.

4 0

3 years ago

What is an exothermic chemical reaction? A. It is a reaction that converts matter to heat. B. It is a reaction that requires hea

nordsb [41]

it is D exo means realease

3 0

3 years ago

Read 2 more answers

What are applications of zeroth law of thermodynamics?

Harrizon [31]

Answer:

Applications of zeroth law of thermodynamics:

1. When we get very hot food, we wait to make it normal. In this case, hot food exchanges heat with surrounding and brings equilibrium.

2. We keep things in the fridge and those things come equilibrium with fridge temperature.

3. Temperature measurement with a thermometer or another device.

4. In the HVAC system, sensors or thermostats are used to indicate temperature. It always comes in a thermal equilibrium with room temperature.

5. If you and the swimming pool you’re in are at the same temperature, no heat is flowing from you to it or from it to you (although the possibility is there). You’re in thermal equilibrium.

6 0

2 years ago

Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat

RSB [31]

Answer:

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

$C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\$

where;

V is the potential difference between the plates

The charge on the plates is given as;

$Q = \frac{V\epsilon _0 A}{d}$

The energy stored in the capacitor is given as;

$E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2$

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

3 0

3 years ago