All categories

4 years ago

Kid A has jumped off the sled! (RIP)

Physics

1 answer:

balandron [24]4 years ago

5 0

Answer:

46.45 m/s

Explanation:

Total momentum before jump = Total momentum after jump

11.7 * ( 36.7 + 45.2 ) = ( 36.7 * (-31.1) ) + ( 45.2 * v )

v*45.2 - 1141.37 = 958.23

v = 2099.6/45.2 = 46.45

You might be interested in

An incident light ray strikes water at an angle of 20 degrees. The index of refraction of air is 1.0003, and the index of refrac

sammy [17]

In order to determine the angle of the refracted ray, we may apply Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is constant for a given wave when it passes through two different media. Mathematically, this is:
n₁sin(∅₁) = n₂sin(∅₂)
Where n is the refractive index. Substituting the values given into the equation:
1.0003 * sin(20°) = 1.33 * sin(∅)
∅ = 14.91

The angle of the refracted ray is 15°.

4 0

3 years ago

Read 2 more answers

A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc

OLga [1]

Answer:

5 N

Explanation:

The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N

7 0

3 years ago

Read 2 more answers

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

3 years ago

Read 2 more answers

A 2.0 g identification reflector glued to one end of a helicopter rotor is spinning at a tangential velocity of 2093 m/s. The re

katrin2010 [14]

372863 juusnjus bhhhanbubhgajhus

7 0

3 years ago

One molecule of bromine (Br2) and two molecules of potassium chloride (KCI) combine in a reaction. How many atoms are in the pro

Liula [17]

In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.

Let's consider the balanced equation for the reaction between 1 molecule of bromine and 2 molecules of potassium chloride. This is a single replacement reaction.

Br₂ + 2 KCl ⇒ 2 KBr + Cl₂

We obtain as products, 2 molecules of potassium bromide and 1 molecule of chlorine.

1 molecule of KBr has 2 atoms, so 2 molecules contribute with 4 atoms.
1 molecule of Cl₂ has 2 atoms.
The 4 atoms from KBr and the 2 atoms from Cl₂ make a total of 6 atoms.

In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.

Learn more: brainly.com/question/21850455

5 0

3 years ago

Read 2 more answers