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saw5 [17]
1 year ago
6

A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards

is the positive velocity direction.
A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,0) and (1,5). This first line is connected to a second line with endpoints (1,5) and (3,-5). This second line is then connected to a third line of endpoints (3,-5) and (6,-5).




A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,0) and (1,5). This first line is connected to a second line with endpoints (1,5) and (3,-5). This second line is then connected to a third line of endpoints (3,-5) and (6,-5).
What is the badger's displacement \Delta xΔxdelta, x from t=2\,\text st=2st, equals, 2, start text, s, end text to 3\,\text s3s3, start text, s, end text?
Physics
1 answer:
Mrrafil [7]1 year ago
5 0

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

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A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?
Talja [164]

Answer:

1.05 N

Explanation:

K = 0.7 N/m

e = 1.5 m

F = ?

from Hooke's law of elasticity

F = Ke

= 0.7×1.5

= 1.05 N

5 0
1 year ago
A worker pushes a 7 kg shipping box along a roller track. Assume friction is small enough to be ignored because of the rollers.
Bingel [31]

Answer:

a) Fₓ = 23.5 N

b) Net force = Fₓ

Explanation:

An image of the question as described is attached to this solution.

From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)

a) From the diagram, the horizontal component of the force is

Fₓ = 25 cos 20° = 23.49 N = 25 N

b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain

Net force = Fₓ - Frictional force

But frictional force is 0 N

Net force = Fₓ

Hope this Helps!!!

6 0
2 years ago
A wire of 1mm diameter and 1m long fixed at one end is stretched by 0.01mm when a lend of 10 kg is attached to its free end.calc
Otrada [13]

Answer:

E = 1.25×10¹³ N/m²

Explanation:

Young's modulus is defined as:

E = stress / strain

E = (F / A) / (dL / L)

E = (F L) / (A dL)

Given:

F = 10 kg × 9.8 m/s² = 98 N

L = 1 m

dL = 10⁻⁵ m

A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²

Solve:

E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)

E = 1.25×10¹³ N/m²

Round as needed.

5 0
2 years ago
A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point
finlep [7]

Answer:

1.17894 m

Explanation:

The rock is at one end of the rod which is 0.211 m from the fulcrum

F = Force

d = Distance

L = Length of rod

M = Mass of rock = 325 kg

g = Acceleration due to gravity = 9.81 m/s²

Torque

\tau=F\times d

Torque of man

\tau_m=F(L-d)\\\Rightarrow \tau_m=695(L-0.211)

Torque of rock

\tau_r=Mg\times d\\\Rightarrow \tau=325\times 9.81\times 0.211\\\Rightarrow \tau=672.72075\ Nm

The torques acting on the system is conserved

\tau_m=\tau_r\\\Rightarrow 695(L-0.211)=672.72075\\\Rightarrow L-0.211=\frac{672.72075}{695}\\\Rightarrow L-0.211=0.96794\\\Rightarrow L=0.96794+0.211\\\Rightarrow L=1.17894\ m

The length of the rod is 1.17894 m

5 0
2 years ago
A pair of slits, separated by 0.150 mm, is illuminated by light having a wavelength of λ = 643 nm. An interference pattern is ob
kumpel [21]

Answer:

0.006 m

Explanation:

d = separation between the slits = 0.150 mm = 0.150 x 10⁻³ m

λ = wavelength of the light = 643 nm = 643 x 10⁻⁹ m

D = distance of the screen where interference pattern is observed = 140 cm = 1.40 m

δ = Path difference

Path difference is given as

δ = Dλ/d

δ = (1.40) (643 x 10⁻⁹)/(0.150 x 10⁻³)

δ = 0.006 m

3 0
2 years ago
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