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Romashka [77]
4 years ago
13

If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v

Physics
1 answer:
dexar [7]4 years ago
5 0
The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units
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What is the average speed of a car traveling for 1000 miles in 6 hours
nlexa [21]

You would divide 6 into 1000 so your answer is B. 166

8 0
3 years ago
A tube of mercury with resistivity 9.84 × 10 -7 Ω ∙ m has an electric field inside the column of mercury of magnitude 23 N/C tha
slava [35]

Answer:

The current through the tube is 73.39A.

Explanation:

The relationship between the resistivity \rho, the electric field E, and the current density J is given by

\rho = \dfrac{E}{J}

This equation can be solved for J to get:

J = \dfrac{E}{\rho}

Since the current is I = J\cdot A

I= J\cdot A  = \dfrac{E}{\rho} \cdot A

Now, for the tube of mercury \rho = 9.84*10^{-7}\: \Omega \cdot m, E = 23N/C, and the area is A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2; therefore,

I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2

\boxed{I = 73.39A.}

Hence, the current through the mercury tube is 73.39A.

5 0
3 years ago
Adam and Natalie push on a cart using 150 N of force.The cart does not move.What is static friction on the cart?
Roman55 [17]

When force is applied on an object the object will not slide

So here object will remain in equilibrium after applying the force

So in this case the condition of equilibrium is maintained due to friction force that is applied on the object.

So here for equilibrium be can say

F_{net} = 0

also we know that

F_{net} = F - F_f

we are given that

F = 150 N ( applied force on the box)

F_{net} = 150 - F_f = 0

F_f = 150 N

So here this friction force on the cart is known as static friction

F_s = 150 N

4 0
4 years ago
Read 2 more answers
Sixty-three joules of heat are added to a closed system. The initial internal energy of the system is 58 J, and the final intern
MakcuM [25]
I posted this because I got it correct.  The answer is 28J.
5 0
3 years ago
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A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to expa
umka2103 [35]

Answer : The correct option is, (B) 279.2 Kpa

Solution : Given,

Initial pressure of gas = 475 Kpa

Initial volume of gas = 4m^3

Final volume of gas = 6.5m^3

Initial temperature of gas = 290 K

Final temperature of gas = 277 K

Using ideal gas equation :

Formula used :

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas

P_2 = final pressure of gas

V_1 = initial volume of gas

V_2 = final volume of gas

T_1 = initial temperature of gas

T_2 = final temperature of gas

Now put all the given values in the above formula, we get the final pressure of the gas.

\frac{(475Kpa)\times (4m^3)}{290K}=\frac{P_2\times (6.5m^3)}{277K}

P_2=279.2Kpa

Therefore, the absolute pressure of the gas after expansion is, 279.2 Kpa

5 0
3 years ago
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