What is the total energy released when 9.11 x10^-31 ki?

Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal pr

Liula [17]

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

$\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}$

therefore : Ra = $\frac{sin120^o * 750}{sin 35^o}$ = 1132 N

To determine the scalar component Rb

$\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}$

therefore : Rb = $\frac{sin 25^o * 750}{sin 35^o}$ = 522.6 N

To determine the orthogonal projection Pa of R onto

Pa = 750 cos $25^o$ = 679.7 N

6 0

2 years ago

The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m

Sonbull [250]

Answer:

density is $10^{6}$ Mg/µL

Explanation:

given data

density of nuclear = $10^{18}$ kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to $10^{6}$ cm³

and here 1 cm³ is equal to 1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = $10^{18}$ kg/m³

density = $10^{18}$ kg/m³ × $\frac{10^{-3} }{10^{6} }$ Mg/µL

density = $10^{6}$ Mg/µL

8 0

2 years ago

Read 2 more answers

When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Sup

Wittaler [7]

Answer:

$W_f = 148.17J$

Explanation:

During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.

Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,

$N=\frac{mg}{2} = \frac{90*9.8}{2} = 441N$

The work done is given by the friction force and the distance traveled,

$W_f = fd = \mu_k Nd$

Where $\mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)$

The thermal energy released through the work done is,

$W_f = 148.17J$

3 0

2 years ago

A 5⁢kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant.

Umnica [9.8K]

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

m = mass of the object.

g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

y = y0 + v0 · t + 1/2 · g · t²

-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

-30 m = 1/2 · g · 4 s²

-30 m = 2 s ² · g

-30 m/2 s² = g

g = -15 m/s²

Then, the magnitude of the gravitational force will be:

F = m · g

F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

8 0

2 years ago

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$