Ask question

Ask question

All categories

3 years ago

7

Pls, help with this! I'm having trouble. will give points!!

1 answer:

seraphim [82]3 years ago

3 0

Answer:

yes

no

no

no

no

yes

Explanation:

You might be interested in

What is an electronic signal

rosijanka [135]

where are the answer choises

6 0

3 years ago

Read 2 more answers

which instrument is best suited for measuring the dimensions of a shoebox? a)a triple-beam balance b)a volumetric flask c)a rule

Aloiza [94]

I believe the correct answer from the choices listed above is option C. The instrument that is <span>best suited for measuring the dimensions of a shoebox would be a ruler. A triple-beam balance is for measuring mass. A volumetric flask is for volume. A caliper is measuring lengths of small objects.</span>

3 0

3 years ago

Read 2 more answers

Four resistors (67 ohm, 83 ohm, 433 ohm, and 309 ohm in that order) are connected in series to a 7.92 V battery of negligible in

Whitepunk [10]

Answer:

.737 v

Explanation:

Since they are in series....they all have the same current running through them.....find the total resistance to calculate the current:

R = 67 + 83 + 433 + 309 = 892 ohm

V/R = current = 7.92 / 892 = 8.87 mAmps

Now the voltage across ecah resistor is I R

for the second one 8.87 ma * 83 ohm = V = .737 V

7 0

2 years ago

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

$v^2-u^2=2as$

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

3 years ago

Because of your success in physics class you are selected for an internship at a prestigious bicycle company in its research and

tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

$M = 1.0 kg$

$r = 0.5 m$

$m = 10 g$

$I = 0.280 kg.m^2$

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

$I = I_{rim} + n * I_{spoke}$

Where n is the number of spokes necessary to construct the wheel.

$I_{rim} = M*r^2 = 1.0 * 0.5^2$

$I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2$

Replacing the values at the general equation we have,

$0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )$

Solving for n,

$n = 36$

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

$M_w= M + n*m$

$M_w = 1.0 + 36* 10 * 10^{-3} Kg$

$M_w = 1.36 Kg$

Therefore the mass of the wheel must be of 1.36Kg

4 0

3 years ago