In a star nuclear fusion occurs in the

Help! I don’t really know what it’s asking

Misha Larkins [42]

You have to do the math of each and see which one adds up to 66.5

6 0

2 years ago

An object is moving at a constant speed along a straight line. Which of the following statements is not true? A. There must be a

Svetradugi [14.3K]

Answer:

False statement = There must be a non-zero net force acting on the object.

Explanation:

An object is moving at a constant speed along a straight line. If the speed is constant then its velocity must be constant. We know that the rate of change of velocity is called acceleration of the object i.e.

$a=\dfrac{dv}{dt}$

a = 0

⇒ The acceleration of the object is zero.

The product of force and acceleration gives the magnitude of force acting on the object i.e.

F = m a = 0

⇒ The net force acting on the object must be zero.

So, the option (a) is not true. This is because the force acting on the object is zero. First option contradicts the fact.

6 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

A 20-kg child is tossed up into the air by her parent. The child is 2 meters off the ground traveling 5 m/s.

nignag [31]

The kinetic energy (KE) is 250 J and the gravitational potential energy (GPE) is 392 J

4 0

3 years ago

Calculate the conductance of a conduit the cross-sectional area of which is 3.0 cm2 and the length of which is 9.0 cm, given tha

pshichka [43]

For resistance we have R=ρ l/a
thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ

σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
=0.26 Ω-1

6 0

3 years ago

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