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Sliva [168]
3 years ago
8

Physics question i appreciate your help please

Physics
2 answers:
Mice21 [21]3 years ago
8 0
The vector c is the answer
Answer choice A
bekas [8.4K]3 years ago
4 0

Really ?  You shouldn't have any trouble locating those two points on the graph.  When you do that, you'll find that the vector ' c ' joins them.  That's choice-A .


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You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the dr
hram777 [196]

Answer:

Explanation:

At a location above the halfway point between the window and the ground

3 0
3 years ago
A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of soun
ella [17]

Answer:

480.2 m

Explanation:

The following data were obtained from the question:

Speed of sound (v) = 343 m/s.

Time (t) = 2.8 s

Distance (x) of the cliff =?

The distance of the cliff from the woman can be obtained as follow:

v = 2x /t

343 = 2x /2.8

Cross multiply

2x = 343 × 2.8

2x = 960.4

Divide both side by the coefficient of x i.e 2

x = 960.4/2

x = 480.2 m

Therefore, the cliff is 480.2 m away from the woman.

4 0
3 years ago
Which describes how the same force affects a small mass and a large mass
n200080 [17]
The same force accelerates a small mass faster than
it accelerates a large mass.

It's easier to get a little red wagon going by pushing it
than it is to get a school bus going by pushing it.
6 0
3 years ago
Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field
kirza4 [7]

Answer: the work done by the force is 0

Explanation:

F (x², xy)

121 = 11²

so R = x² + y² = 11²

p = x². Q = xy

Δp/Δy = 0, ΔQ/Δx

using Green's theorem

woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA

=  (x² + y² = 121)_∫∫ yΔA

now let x = rcosФ, y = rsinФ

ΔA = rΔrΔФ

so r from 0 to 11

and Ф from 0 to 2π

= 0_∫^2π   0_∫^11  rsinФ × rΔrΔФ

= 0_∫^2π SinФΔФ   0_∫^11  r²Δr

= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0

therefore the work done by the force is 0

3 0
4 years ago
wide tube that extends down from the bag of solution, which hangs from a pole so that the fluid level is 90.0 cm above the needl
Bingel [31]

Answer:

The average gauge pressure inside the vein is 110270.58 Pa

Explanation:

This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa\\

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa

6 0
3 years ago
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