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3 years ago

13

What is the effective resistance of a car’s starter motor when 150 A flows through it as the car battery applies 12.0 V to the m

otor?

1 answer:

QveST [7]3 years ago

8 0

Answer:

From ohms law,

V=IR

R=V/I =12.0/150 =0.08 ohm.

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a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago

A flap of tissue that prevents blood from flowing back is a(n)________ .

ioda

A valve is a flap of tissue that prevents blood from flowing back

6 0

3 years ago

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

5 0

3 years ago

When is a hypothesis developed in the scientific method?

ivanzaharov [21]

Answer:

aren't there pics ?

Explanation:

7 0

3 years ago

Read 2 more answers

4) A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

VMariaS [17]

Answer:

(a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

Explanation:

Given that,

A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Suppose, the distance between 40 yard line and 25 yard line is 20 yard.

(a). We need to calculate their speed during that run

Using formula of speed

$v=\dfrac{d}{t}$

Where. d = distance

t = time

Put the value into the formula

$v=\dfrac{18.288}{2}$

$v=10\ m/s$ during that run

(b). We need to calculate their velocity

Using formula of speed

$v=\dfrac{\Delta d}{\Delta t}$

Put the value into the formula

$v=\dfrac{22.86-36.58}{2}$

$v=-6.86\ m/s$

Negative sign shows the direction of motion.

(c). If they kept running at that velocity for another 1.3 seconds,

We need to calculate the final position

Using formula of position

$d=vt$

Put the value into the formula

$d=6.86\times1.3$

$d=8.91\ m$

Hence, (a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

8 0

3 years ago