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MaRussiya [10]
3 years ago
8

If the body temperature of a person is recorded as 98.6°F, what is the person’s body temperature in °C? 

Physics
2 answers:
blagie [28]3 years ago
6 0
98.6 degrees F is 37 degrees C
Greeley [361]3 years ago
6 0
37.0<span>°C is the temperature in celcius</span>
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Una patinadora de 50 kg parte del reposo y después de recorrer 3k alcanza una velocidad de 15 m/s. ¿Qué fuerza neta experimenta
e-lub [12.9K]

Answer:

F_{net} = 1.875\,N

Explanation:

Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:

F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]

F_{net} = 1.875\,N

6 0
3 years ago
I need help on 6, 7, 8, and 9
Ivahew [28]
6 is b. part B on 6 is a. 7 is a. partB ON 7 b
5 0
3 years ago
An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of
GaryK [48]

Explanation:

The given data is as follows.

     Velocity of bullet, c_{p} = 814.8 m/s

    Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

               t = \frac{24.7}{343}      (velocity in air = 343 m/s)

                 = 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

               d_{b} = c_{b} \times t

                          = 814.8 \times 0.072

                          = 58.66

                          = 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0
3 years ago
The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
3 years ago
: Do you attract the Earth or the Earth attracts you which one is attracting with a larger force?
seraphim [82]

Answer: The Earth and you are attracted to the centers of each other by a pair of equal gravitational forces. The size of the force attracting you toward the center of the Earth is your "weight" on Earth. The size of the force attracting the Earth toward the center of you is the Earth's "weight" on you.

Explanation:

8 0
3 years ago
Read 2 more answers
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