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pentagon [3]
3 years ago
14

The image above shows two opposite forces acting on a rope, what can we say is true about the affect of the forces on

Physics
2 answers:
tino4ka555 [31]3 years ago
7 0

Answer:

your answer is D

Explanation:

..........

WITCHER [35]3 years ago
7 0

Answer:

d

Explanation:

UsaTestPrep

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Alien A lifts a 500-newton child from the floor to a height of 0.40 meters in 2 seconds
vivado [14]
Strong alien you got there good luck bud you never asked a question
4 0
3 years ago
The dimensions for force are the product of mass and length divided by time squared. Newton's second law states that force equal
maw [93]

Answer:

L/T^2.

Explanation:

Dimensional analysis is defined as the analysis of the relationships between different physical quantities by identifying their base quantities (such as length, mass, time, temperature etc.) and units of measure (such as miles vs. kilometers, pounds vs. kilograms etc.).

F = m * a

kg(m/s^2) = kg * a

In dimensions,

a = L/T^2.

5 0
3 years ago
The picture shows a loop of wire rotating in a magnetic field in a generator. Determine the direction of the current through the
BaLLatris [955]
I believe the answer is b
3 0
2 years ago
Read 2 more answers
Vector → A points due north and vector → B points due west. Both vectors have the same magnitude. Which of the following vectors
-BARSIC- [3]

Answer:

B - A

Explanation:

For the combination of 2 vector to due southwest, 1 vector must due south and the other vector due west. Since vector B is already due west, vector A should due south. As vector A is already due north, vector -A would due south. So the combination of B + (-A) or B - A should points southwest

5 0
2 years ago
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre
Alisiya [41]

Answer:

0.546 \hat k

Explanation:

From the given information:

The force on a given current-carrying conductor is:

F = I ( \L  \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})

where the length usually in negative (x) direction can be computed as

\L ^ {\to }  = -x\hat i \\dL\limits ^ {\to }- dx\hat i

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})

F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)

F = I \int^3_1  - 9.0x^2 \ dx \hat k

F = I  (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k

F = I  (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ]  \hat k

where;

current I = 7.0 A

F = (7.0 \ A)  (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ]  \hat k

F = (7.0 \ A)  (9.0) \bigg [\dfrac{26}{3} \bigg ]  \hat k

F = 546 × 10⁻³ T/mT \hat k

F = 0.546 \hat k

4 0
3 years ago
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