All categories

4 years ago

Express the number in scientific notation: -8,675,300.0

Physics

1 answer:

stich3 [128]4 years ago

4 0

Answer:

It would be -8.6753 x 10 to the 6th power

Explanation

You might be interested in

I dont understand??????????????

Strike441 [17]

The sum of the two vectors in A, B, and C is equal to the sum of the two vectors above the line. The sum of the two vectors in D isn't.

7 0

3 years ago

A physicist is calibrating a spectrometer that uses a diffraction grating to separate light in order of increasing wavelength (λ

cluponka [151]

Answer:

Explanation:

Given

$N=3680 cm^{-1}$

therefore slit spacing $d=\frac{1}{N}=\frac{1}{3680}=2.717\times 10^{-4}\ cm$

since $d\sin \theta =n\lambda$

for $n=1$

$d\sin \theta =\lambda$

Now,at $\theta _1=12.9^{\circ},\Rightarrow \lambda _1=6.0657\times 10^{-7}\ m=606.57\ nm$

at $\theta _2=14.2^{\circ}\Rightarrow \lambda_2=666.5\ nm$

at $\theta _3=15^{\circ}\Rightarrow \lambda_3=703.21\ nm$

5 0

3 years ago

A ranger in a national park is driving at 52 km/h when a deer jumps onto the road 87 m ahead of the vehicle. After a reaction ti

lys-0071 [83]

Answer:

Time, t = 0.23 seconds

Explanation:

It is given that,

Initial speed of the ranger, u = 52 km/h = 14.44 m/s

Final speed of the ranger, v = 0 (as brakes are applied)

Acceleration of the ranger, $a=-4\ m/s^2$

Distance between deer and the vehicle, d = 87 m

Let d' is the distance covered by the deer so that it comes top rest. So,

$d'=\dfrac{v^2-u^2}{2a}$

$d'=\dfrac{-(14.44)^2}{2\times -4}$

d' = 26.06 m

Distance between the point where the deer stops and the vehicle is :

D=d-d'

D=87 - 26.06 = 60.94 m

Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :

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

$t=\dfrac{14.44}{60.94}$

t = 0.23 seconds

Hence, this is the required solution.

4 0

3 years ago

Obtain a volume of 12.5 millimeters of liquid in the diagram

tiny-mole [99]

Answer:

D remove 1.5 ML of liquid.

Explanation:

5 0

3 years ago

Read 2 more answers

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

$f=\mu mg$

Put the value into the formula

$f=0.8\times37$

$f=29.6\ N$

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

$f = \mu N$

Put the value in to the formula

$f=0.4\times37$

$f=14.8\ N$

Hence, The frictional force acting on the block is 14.8 N.

6 0

3 years ago