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2 years ago

15

A man is pulling on his dog with a force of 34.7 N directed at an angle of 24.4° to the horizontal. Find the x component of this

force.

1 answer:

stepladder [879]2 years ago

7 0

Answer:

$31.7\:\mathrm{N}$

Explanation:

We can use basic trig in a right triangle to solve this problem. Let the x component of the force be $x$ . We have the following equation:

$\cos 24^{\circ}=\frac{x}{34.7},\\x=34.7\cos 24^{\circ}\approx \boxed{31.7}$

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DNA’s diameter is 0.000000002 meters. What is this measurement in scientific notation? a. 2.0 x 109 m

NARA [144]

Answer:

0.000000002 m=2.0*10⁻⁹ m

Explanation:

Scientific notation allows us to write very large or very small numbers in abbreviated form. This notation simply consists of multiplying by a power of base 10 with a positive or negative exponent.

A number written in scientific notation has the form:

a*10ⁿ

where:

the coefficient a has a value such that 1 ≤ a <10
n is an integer. Represents the number of times the decimal point is shifted. It is always a whole number, positive if it is shifted to the left, negative if it is shifted to the right.

So to write the number 0.000000002 in scientific notation, the following steps are performed:

The decimal point is moved to the right as many spaces until it reaches the right of the first digit.

This number is then written, which will be the coefficient a in the expression of the previous product. So a=2.0
The base 10 is written with the exponent equal to the number of spaces that the comma moves. So n=9. But this is a negative number because the comma shifts to the right.

So, you get: <u><em>0.000000002 m=2.0*10⁻⁹ m</em></u>

3 0

3 years ago

A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t

sertanlavr [38]

Explanation:

The given data is as follows.

C = $20 \times 10^{-6} F$

R = $100 \times 10^{3}$ ohm

$Q_{o} = 100 \times 10^{-6}$ C

Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

$Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

$13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

0.135 = $e^{\frac{-t}{2}}$

$e^{\frac{t}{2}} = \frac{1}{0.135}$

= 7.407

$\frac{t}{2} = ln (7.407)$

t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0

2 years ago

The transfer of energy by the movement of particles that are in contact with each other

jeka57 [31]

Answer: I belive that the Answer is C.) Conduction

Explanation:

3 0

3 years ago

Read 2 more answers

Two dogs fight over a bone. The larger of the two pulls on the bone to the right with a force of 42 N. The smaller one pulls to

tamaranim1 [39]

Answer:

Explanation:

A

35 N Small Dog <=======BONE=========> Bigger Dog 42 N

B

Fnet = Large Dog - small dog The forces are subtracted because they are acting in opposite Directions.

Fnet = 42 - 35

Fnet = 7 N

C

m = 2.5 kg

F = 7 N

a = ?

F = m * a

7 = 2.5 a

a = 7 / 2.5

a = 2.8 m/s^2

4 0

1 year ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago