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3 years ago

What kind of radiation is emitted in the following nuclear reaction

Physics

1 answer:

dedylja [7]3 years ago

7 0

Answer:

The answer is B. (beta).

Explanation:

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The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.

klemol [59]

The resistance is 27.5 ohms

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3 years ago

Read 2 more answers

Is there a "real" simple machine that has an efficiency of 100%?

VARVARA [1.3K]

Answer:

Explanation:

No, a machine cannot be 100% efficient. This is due to the movement of the moving parts siding against each other and causing friction. This friction is the one that creates heat and causes wear and tear between moving ports f the machine hence making the machine to decrease in efficiency with time

8 0

3 years ago

Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa

Serga [27]

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

$v_r$ = Speed of river

$v_c$ = Speed of canoe

$v_r+v_c=2.8\ m/s$

$v_r-v_c=-1.9\ m/s$

Adding the equations we get

$2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s$

$0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s$

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

6 0

3 years ago

In what state of matter are particles vibrating and moving at high speeds?

VARVARA [1.3K]

Answer:

Gas

Explanation:

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