Ask question

Ask question

All categories

2 years ago

6

a falling skydiver slows from a speed of 52 m/s to 8 m/s in 0.8 sec as the parachute opens. what is the diver's acceleration and

the displacement ?

1 answer:

Cloud [144]2 years ago

8 0

I found the answers here. Hope this helps you! https://1.cdn.edl.io/sJTle6yxt3qVq7jHfdHRZJ3Xogj7ps6swBO9umNcZ6PO3SMN.docx

You might be interested in

While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properti

Ede4ka [16]

Answer: There are many possible elements, and they are all in the same vertical column as bromine.

Explanation:

In a periodic table, the elements are arranged according to the atomic number. The elements arranged in the same vertical column (known as groups) have same valence configuration and therefore have same chemical properties. Hence, there would be more possible elements having same chemical properties in the same vertical column (group) as Bromine.

7 0

2 years ago

Read 2 more answers

A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the fi

allochka39001 [22]

Answer:

a. q2 = 16.4μC, positive charge

b. F = 0.900N

c. downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

$F_e=k\frac{q_1q_2}{r^2}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

$q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C$

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

3 0

2 years ago

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l

Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

$u=+6.3 m/s$ is the initial velocity (positive because it is upward)

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

$v=+6.3 + (-9.8)(1.7)=-10.4 m/s$

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0

2 years ago

Read 2 more answers

Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water - an impressive feat. Suppose

dmitriy555 [2]

Answer:14 m

Explanation:

Given

Vertical jump make by the dolphin is given by $h=7\ m$

Suppose the dolphin jump with an initial velocity of $u$

so u is given by $u^2=2\cdot g\cdot h$

If dolphin launches at an angle $\theta$ then maximum horizontal range is given by

assuming the of Dolphin to be Projectile so range is given by

$R=\frac{u^2\sin 2\theta }{g}$

substitute the value of $u^2$

$R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}$

$R=2h\sin 2\theta$

Range will be maximum for $\theta =45^{\circ}$

thus $R_{max}=2\times 7\times 1=14\ m$

3 0

2 years ago

Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield

ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

Explanation:

Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be

Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

mP = 100 kg* 0.037/0.079 = 46.835 kg

then the yield YP of the product is

YP= mP / 100 kg = 46.835 kg / 100 kg = 46.835 %

YP= 46.835 %

the concentration of solids Cs is

Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs

Cs = 100 kg* 0.128 / 46.835 kg = 0.2733 = 27.33%

Cs = 27.33%

3 0

2 years ago