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Charra
14 days ago
11

Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0

.7 km in the original direction. Let Tiana's original direction be the positive direction. What are the displacement and distance she jogged?
Physics
1 answer:
inna [2.2K]14 days ago
7 0

Answer:

Distance: 4.6 km Displacement= -0.2 km

Explanation:

The overall distance covered: 1.5 + 2.4 + 0.7 = 4.6 km

Displacement calculation: 1.5 - 2.4 + 0.7 = -0.2 km

The displacement could also simply be stated as 0.2 km depending on whether negative value is preferred.

You might be interested in
If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight t
Sav [2226]

Para calcular el peso utilizamos la fórmula:
F=m*g=882N

Cuando tenemos dos cuerpos, empleamos la fórmula de la gravedad general:
F=G* \frac{ M_{p}*M_{E} }{ r^{2} }
Donde:
G = constante de gravedad = 6.67* 10^{-11} m^{3} kg^{-1} s^{-2}
Mp = masa de la persona = 882 / 9.81= 89.9kg
ME = masa de la Tierra = 5.97* 10^{24} kg
r = distancia entre la Tierra y la persona

A partir de estas dos fórmulas, deducimos que los lados izquierdos son equivalentes, por lo tanto, los lados derechos también deben serlo.
m*g=G* \frac{ M_{p}*M_{E} }{ r^{2} }

Resolviendo esta ecuación para r:
r= \sqrt{G* \frac{ M_{p}*M_{E}}{M_{p}*g}
r=6371116m = 6371.116km

Esta es la distancia desde el centro de la Tierra. El radio de la Tierra es 6370km y la altura sobre la superficie es 6371.116 - 6370 = 1.116km o 1116m.

3 0
22 days ago
Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.
inna [2205]
The heat required to raise the temperature of a substance by \Delta T is represented by
Q=m C_p \Delta T
where m stands for the mass of the substance and C_p indicates the specific heat of the substance. In this situation, we possess m=100~g=0.1~Kg and C_p=4.19~KJ/(Kg K), the specific heat of water.
Consequently, we can ascertain the temperature rise \Delta T:
\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C
Initially, the water's temperature was 25^{\circ}C, so the end temperature should be
T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ
where C_L denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of 100^{\circ}C (the temperature at which the phase change begins)

4 0
25 days ago
A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
ValentinkaMS [2425]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is \frac{4}{3} \pi (\frac{D}{2})^3

Thus \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

The radius of the disk is 1.15(\frac{ D}{2} )

Applying angular momentum conservation;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disk = m*\frac{ \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
25 days ago
Incoming solar radiation (light energy) is absorbed by the Earth's surface and converted to outgoing infrared radiation (heat en
Softa [2029]
An increase in temperature (Global warming) is observed. The solar radiation is transformed into heat energy absorbed by Earth's surface. In line with the law of conservation of energy, energy can only transition forms rather than disappear. If an increasing quantity of energy accumulates on Earth with minimal release, this imbalance in energy demand leads to a rise in temperature due to excessive heat absorption, largely a result of pollution from fossil fuel combustion releasing CO2 and other harmful emissions. Ordinarily, the residual solar energy would escape back into space, but CO2 and similar contaminants trap this heat, thus elevating Earth's temperature.
8 0
11 days ago
A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At
kicyunya [2264]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + \frac{1}{2}at^{2}

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + \frac{1}{2}(2.25 x 15.4^{2})

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

v^{2} = u^{2} + 2gs

0^{2} = 34.65^{2} + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying v^{2} = u^{2} + 2gs

v^{2} = 0^{2} + 2(9.81 x 328)

v^{2} = 0 + 6435.36

v = \sqrt{6435.36} = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0
25 days ago
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