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Paraphin [41]
3 years ago
6

Question 15 of 15

Physics
1 answer:
castortr0y [4]3 years ago
7 0

Answer:

a and b

Explanation:

are the correct answers

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Joey drives his Skidoo 13 kilometres north. He stops for lunch and then drives 10kilometres south. What distance did he cover? W
olganol [36]

Answer:

Total distance covered (scalar quantity) = 23 km

Displacement (vector quantity) = 3 km north from the original starting point

Explanation:

Since he drove 13 km north and then 10 km south, the total distance he cover in his drive was: 13 km + 10 km = 23 km

On the other hand, his displacement was 3 km north from where he started.

7 0
3 years ago
Why does a charged balloon stick to a neutral wall?
Jobisdone [24]
<span>When you bring a charged object, such as your balloon, near a neutral object that is classified as an insulator, than a temporary charge is induced in the neutral object. If the charged object is positive, then electrons in the neutral object will be attracted toward the charged object, creating a temporary imbalance of charges in the neutral object.</span>



7 0
3 years ago
A student puts a beaker of boiling water where it touches a block of ice. Which two statements are true? A. Thermal energy will
rodikova [14]

Answer: C and D

Explanation: I’m pretty sure it’s C and D I got the same question but cold water let me know please, thanks!

6 0
4 years ago
Read 2 more answers
The liquid in the cup is brown" is an example of an ..............
erastova [34]
Eaither D or A bit I am leaning more towards D
3 0
3 years ago
Read 2 more answers
Se deja caer una pelota inicialmente en reposo desde una altura de 50m sobre el nivel del suelo. ¿cuanto tiempo requiere para ll
umka2103 [35]

Answer:

a) t = 3.2 s

b) v_{f} = -32 m/s

Explanation:

a) El tiempo requerido para llegar al suelo se puede calcular usando la siguiente fórmula:

t = \sqrt{\frac{2y_{0}}{g}}

En donde:

y_{0}: es la altura inicial = 50 m

g: es la gravedad = 10 m/s²

t = \sqrt{\frac{2y_{0}}{g}} = \sqrt{\frac{2*50 m}{10 m/s^{2}}} = 3.2 s

Entonces, el tiempo requerido para llegar al suelo es 3.2 s.

b) La rapidez de la pelota justo antes del choque es el siguiente:

v_{f} = v_{0} - gt

En donde:

v_{0}: es la velocidad inicial = 0 (dado que se deja caer en resposo)

v_{f} = v_{0} - gt = 0 - 10 m/s^{2}*3.2 s = -32 m/s

Por lo tanto, la rapidez de la pelota justo en el momento anterior del choque es -32 m/s (el signo negativo es porque la pelota está cayendo).

Espero que te sea de utilidad!

6 0
3 years ago
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