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dalvyx [7]
3 years ago
5

100 degrees C equals how many degrees F?​

Physics
1 answer:
elena-s [515]3 years ago
7 0

Answer:

212 f

Explanation:

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What will happen to force?
telo118 [61]

Answer:

it will double because im right

8 0
3 years ago
Recall specific heat of water is 4186 j/kg/C. Find the specific heat of sample.
Paraphin [41]

Answer:

Shown by explanation;

Explanation:

The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)

Assumption;I assume the mass of the samples are : 109g and 192g

∆T= 30.1-21=8.9°c.

The heat of the samples are for 109g are:

0.109 × 4186 × 8.9 =4060.84J

For 0.192g are;

∆T= 67-30.1-=36.9°c

0.192 × 4186×36.9=29656.97J

5 0
3 years ago
Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10
gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

\Sigma \tau=0

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

-\tau_l+\tau_p+\tau_{cm}=0          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

\tau=Fd           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N

The force applied by the left support is 2123.33 N

8 0
3 years ago
Near the equator, the Earth's magnetic field points almost horizontally to the north and has magnitude B=.5 x 10^-4T. What shoul
Nataly [62]
Base in your question about the magnetic field of the Earth near the equator where as its almost horizontally to the north and has magnitude of B=0.5x10^-4t, the answer is <span>Velocity of electron will be westwards.</span>
6 0
3 years ago
An automobile having a mass of 2000 kg deflects its suspension springs 0.02 m under static conditions. Determine the nafural fre
alexandr1967 [171]

Answer:Frequency = 3.525 Hertz

Explanation:In static equilibrium, kd =mg

Where k= effective spring constant of the spring.

mg= The weight of the car.

d= static deflection.

Therefore, w =SQRTg/d

w = SQRT 9.81/0.02

w= 22.15 rad/sec

Converting to Hertz unit for frequency

1 rad/s = 0.1591

22.15rad/s=?

22.15 × 0.1591= 3.525 hertz

7 0
3 years ago
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