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2 years ago

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A ball is thrown vertically upwards with a velocity of 9.8 m/s. Its velocity after 1 second will be

1 answer:

Reil [10]2 years ago

5 0

$\LARGE{ \underline{\underline{ \purple{ \bf{Required \: answer:}}}}}$

GiveN:

Initial velocity = 9.8 m/s²
Accleration due to gravity = -9.8 m/s²
Time taken = 1 s

To FinD:

Final velocity of the ball?

Step-by-step Explanation:

Using the first Equation of motion,

⇒ v = u + gt

⇒ v = 9.8 + -9.8(1)

⇒ v = 0 m/s

The final velocity is hence <u>0</u><u> </u><u>m</u><u>/</u><u>s</u><u>.</u>

<h3>Note:</h3>

While solving questions of under gravity motions using equations of motion, remember the sign convection to avoid mistakes.
You can consider positive above the ground and negative for towards it.

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3 years ago

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

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2 years ago

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3 years ago

Read 2 more answers

What molecular geometry would be expected for f2 and hf?

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The molecular geometry of both F2 and HF is linear.

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2 years ago

Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo

lisabon 2012 [21]

Answer:

El peso del cartel es 397,97 N

Explanation:

La tensión dada en cada segmento del cable = 2000 N

El desplazamiento vertical del cable = 50 cm = 0,5 m

La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

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3 years ago