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3 years ago

True or False. If your skin is wet, the body's resistance to electric shocks increases.

Physics

2 answers:

Svetllana [295]3 years ago

4 0

Answer:

Pretty sure its false, water increases electric shock

Explanation:

Tpy6a [65]3 years ago

4 0

Answer: False

Explanation:

It would never increase because you are already wet, and you can not get any wetter.

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A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a

VashaNatasha [74]

Answer:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

$r_1 = 45 sin15 \hat i + 45 cos15 \hat j$

$r_1 = 11.64 \hat i + 43.46\hat j$

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

$r_2 = 30 cos15\hat i + 30 sin15 \hat j$

$r_2 = 28.98\hat i + 7.76 \hat j$

now the displacement of the boat is given as

$d = r_2 - r_1$

$d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)$

$d = 17.34 \hat i - 35.7 \hat j$

so the magnitude is given as

$d = \sqrt{17.34^2 + 35.7^2}$

$d = 39.7 km$

4 0

3 years ago

Help pleaseeeeeeeeeeeeee

kap26 [50]

The frictional force is in the opposite direction

7 0

3 years ago

A clay vase on a potter's wheel experiences an angular acceleration of 5.69 rad/s2 due to the application of a 16.0-n m net torq

Digiron [165]

The equivalent of the Newton's second law for rotational motions is:
$\tau = I \alpha$
where
$\tau$ is the net torque acting on the object
$I$ is its moment of inertia
$\alpha$ is the angular acceleration of the object.

Re-arranging the formula, we get
$I= \frac{\tau}{\alpha}$
and since we know the net torque acting on the (vase+potter's wheel) system, $\tau=16.0 Nm$ , and its angular acceleration, $\alpha = 5.69 rad/s^2$ , we can calculate the moment of inertia of the system:
$I= \frac{\tau}{\alpha}= \frac{16.0 Nm}{5.69 rad/s^2} =2.81 kg m^2$

8 0

3 years ago

What’s the atomic number for magnesium

zmey [24]

Answer:

Explanation:

Symbol: Mg

Atomic mass: 24.305 u

Atomic number: 12

Density: 1.738 g/cm³

5 0

3 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: