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victus00 [196]
3 years ago
5

More pts first 2 so hurry up

Physics
2 answers:
Norma-Jean [14]3 years ago
8 0

Answer:

oh 50 points! how did you do it??!?!?!?! I see up to 8 points only

maria [59]3 years ago
4 0

Answer:

Hello, thank you for giving out points, you are very kind !

You might be interested in
A wooden block meauring 40cm x 10cm x 5cm has a mass 850gm . find the density of wood?
Lelechka [254]

Answer:

Explanation:

Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3

5 0
3 years ago
Two cars leave towns 680 kilometers apart at the same time and travel toward each other. one car's rate is 10 kilometers per hou
Evgen [1.6K]

D = distance between the cars at the start of time = 680 km

v₁ = speed of one car

v₂ = speed of other car = v₁ - 10

t = time taken to meet = 4 h

distance traveled by one car in time "t" + distance traveled by other car in time "t" = D

v₁ t + v₂ t = D

(v₁ + v₂) t = D

inserting the values

(v₁ + v₁ - 10) (4) = 680

v₁ = 90 km/h

rate of slower car is given as

v₂ = v₁ - 10

v₂ = 90 - 10 = 80 km/h

5 0
4 years ago
A Nichrome wire 44 cm long and 0.30 mm in diameter is connected to a 3.1 V flashlight battery. What is the electric field inside
Alexeev081 [22]

Answer:

7.05 Volts/m

Explanation:

L = length of the Nichrome wire = 44 cm = 0.44 m

V = Potential difference across the end of the wire = battery voltage = 3.1 Volts

E = magnitude of electric field inside the wire

Magnitude of electric field inside the wire is given as

E = \frac{V}{L}

Inserting the values

E = \frac{3.1}{0.44}

E = 7.05 Volts/m

4 0
3 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
The chocolate in your favorite candy bar has the chemical formula: C7H8N4O2. What elements are present in this formula? Carbon,
fenix001 [56]

Carbon, hydrogen, nitrogen and oxygen are all non metals.

5 0
3 years ago
Read 2 more answers
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