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Debora [2.8K]
3 years ago
7

Kid A has jumped off the sled! (RIP)

Physics
1 answer:
balandron [24]3 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

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3.9 divided by 15 gcuu
Vsevolod [243]

0.26 there you go buddy

8 0
3 years ago
Read 2 more answers
Two point charges, 1.8 pC and −1.8 pC, are separated by 7 µm. What is the dipole moment of this pair of charges?
Salsk061 [2.6K]

Answer: 12,600,000Cm

Explanation:

From the data's;

Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C

Distance = 7 micrometer, is equal to 0.0000070m

From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.

Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm

NB: The charges are identical.

3 0
2 years ago
How does a 12-month lunar calendar differ from our 12-month solar calendar?.
horrorfan [7]

Answer:

It has about 11 fewer days. It does not have seasons. Its new year always occurs in February instead of on January 1.

Explanation:

8 0
2 years ago
a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0
3 years ago
Two parallel wires are separated by 5.60 cm, each carrying 2.65 A of current in the same direction. (a) What is the magnitude of
ololo11 [35]

Explanation:

It is given that,

The separation between two parallel wires, r = 5.6 cm = 0.056 m

Current in both the wires is 2.65 A

(a) We need to find the magnitude of the force per unit length between the wires. It can be given by :

\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.65\times 2.65}{2\pi \times 0.056}\\\\\dfrac{F}{l}=2.5\times 10^{-5}\ N/m

(b) As the current is in same direction, the wires will attract each other.

5 0
3 years ago
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