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3 years ago

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What two products are formed from the fission of uranium?

1 answer:

zheka24 [161]3 years ago

3 0

Fragments and gamma<span> rays</span>

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Your portable phone's 12V battery is dead. It's a holiday and all the stores are closed. You do have two 1.5V batteries from you

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Answer:

No.

Explanation:

A transformer requires AC to work. A battery delivers DC only.

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4 years ago

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What kind of cell is in a dragonfly? A eukaryotic or prokaryotic?

ICE Princess25 [194]

A dragonfly is a prokaryotic cell

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3 years ago

A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on t

Phoenix [80]

Answer:

a. $I=981.34 N*s$

b. $v_f=3.96 m/s$

c. $v_{f1}=3.63m/s$

d. $y_f=0.673m$

Explanation:

Given: $m=67kg$ , $h=0.720m$ , $0$

a.

$I=\int\limits^{t_1}_{t_2} {F(t)} \, dt$

$F(t)=9200*t-11500t^2$

$I=\int\limits^{0.8s}_{0s}{9200*t-11500*t^2} \, dt$

$I=4600*t^2-3833.3*t^3|(0.80,0)$

$I=2944-1962.66=981.35$

$I=981.34 N*s$

b.

$v_f^2=v_i^2+a*y'$

Starting from the rest

$v_f^2=0+2*9.8m/s^2*0.80s$

$v_f^2=15.68$

$v_f=\sqrt{15.68m^2/s^2}=3.96 m/s$

c.

$I_{total}=p_f$

$I_1-m*g*d=m*v_{f1}-m*v_f$

$981.34-67kg*9.8m/s^2*0.720=67.0kg*v_{f1}-67.0kg*(-3.96m/s)$

Solve to vf

$v_{f1}=3.63m/s$

d.

$v_f^2=v_i^2+2*a*y_f'$

$y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2$

$y_f=0.673m$

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3 years ago

Which of the following would reduce the resistance of a metal wire?

Lady_Fox [76]

All of the above can affect the resistance of a metal

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4 years ago

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A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea

Savatey [412]

Answer:

$41.81^{\circ}$

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector $\overrightarrow{OA}$ is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, $\vec {R}$ must be in the north direction.

Let $\overrightarrow{AB}$ is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector $\overrightarrow {OA}$ and head of the vector $\overrightarrow{AB}$ must lie on the north-south line.

Now, for this condition, from the triangle OAB

$|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|$

$\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3$

$\Rightarrow \theta=\sin^{-1}\frac23$

$\Rightarrow \theta=41.81^{\circ}$

Hence, the kayaker must paddle in the direction of $41.81^{\circ}$ in the north of east direction.

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4 years ago