All categories

3 years ago

2. How long will it take for a ball thrown vertically upward with an initial velocity of

Physics

1 answer:

KengaRu [80]3 years ago

6 0

Answer:

13.33 seconds

Explanation:

At maximum height, the equation of motion becomes:

v = u + at

Since the object was thrown vertically, the initial velocity (u) is zero and the acceleration (a) becomes the acceleration due to gravity (10 m/s2). The equation becomes:

v = at

v = 480 km/hr = 133.333 m/s

10t = 133.333

t = 133.333/10

t = 13.33 seconds.

The time for the ball thrown vertically with a velocity of 480 km/hr to reach the maximum height is 13.33 seconds.

You might be interested in

A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The

malfutka [58]

Answer:

$\theta = 23.32^o$

$\mu_s = 0.27$

$s = 0.948 \ m$

Explanation:

From the question we are told that

The mass of the bag is $m_b = 25.0 \ kg$

The normal force experienced is $F_n = 225 \ N$

The maximum acceleration of the bag is $a = 2.40 \ m/s^2$

Generally this normal force experience by the bag is mathematically represented as

$F_n = mg cos \theta$

=> $225 = (25 * 9.8) cos \theta$

=> $0.9183 = cos \theta$

=> $\theta = cos^{-1}[0.9183]$

=> $\theta = 23.32^o$

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

$F_f = F_s$

Hence $F_f$ is mathematically represented as

$F_f = \mu_s * F_n$

While $F_s$ is mathematically represented as

$F_s = m * a$

$\mu_s * F_n = m * a$

=> $\mu_s * 225 = 25 * 2.40$

=> $\mu_s = 0.27$

Generally from the workdone equation we have that

$KE_f - KE_i = W_f$

Here $W_f$ is the work done by friction which is mathematically represented as

$W_f = m * g * \mu_k * s$

Here s is the distance covered by the bag

$KE_f$ is zero given that velocity at rest is zero

and

$KE_i = \frac{1}{2} * m* v_i^2$

$\frac{1}{2} * m* v_i^2 = m * g * \mu_k * s$

=> $\frac{1}{2} * v_i^2 = g * \mu_k * s$

substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that

$\frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s$

=> $s = 0.948 \ m$

4 0

3 years ago

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angl

vazorg [7]

Answer:879.29 N-m

Explanation:

Given

mass of first child $m_1=44 kg$

distance of first child from tree is $r_1=1 m$

tree is inclined at an angle of $\theta =27^{\circ}$

mass of second child $m_1=27 kg$

distance of second child from tree is $r_2=2.1 m$

Weight of first child $=m_1g=431.2 kg$

Weight of second child $=m_2g=264.6 kg$

Torque of first child weight $=m_1g\cos \theta \cdot r_1$

$T_1=44\times 9.8\times \cos 27\times 1=384.202 N-m$

Torque of second child weight $=m_2g\cos \theta \cdot r_2$

$T_2=27\times 9.8\times \cos 27\times 2.1=495.096 N-m$

Net torque $T_{net}=T_1+T_2=384.202+495.096=879.29 N-m$

6 0

3 years ago

Please help on this one?

Maru [420]

The correct answer for this problem is c

3 0

3 years ago

Read 2 more answers

In preparation for the final exam, our astronomy study group has reconvened to discuss Mercury's unique orbital properties. They

grigory [225]

Answer:

The only incorrect statement is from student B

Explanation:

The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.

Let's examine student claims using these rotation periods

Student A. The time for 4 turns around the sun is

t = 4 88

t = 352 / 58.7 Earth days

In this time I make as many rotations on itself each one with a time to = 58.7 Earth days

#_rotaciones = t / to

#_rotations = 352 / 58.7

#_rotations = 6

therefore this statement is TRUE

student B. the planet rotates 6 times around the Sun

t = 6 88

t = 528 s

The number of rotations on itself is

#_rotaciones = t / to

#_rotations = 528 / 58.7

#_rotations = 9

False, turn 9 times

Student C. 8 turns around the sun

t = 8 88

t = 704 days

the number of turns on itself is

#_rotaciones = t / to

#_rotations = 704 / 58.7

#_rotations = 12

True

The only incorrect statement is from student B

6 0

2 years ago

Order the speed of sound through these materials from the slowest to the fastest.

Sholpan [36]

Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

Step 1:

Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

Speed of sound in warm air is approximately 338.89 meters per second

Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

In warm air sound travels faster than that of sound travelling nature in cold air.

∴ Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Speed of sound in cold air the slowest while Speed of sound in water is the fastest mean.

8 0

3 years ago