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3 years ago

PLEASE HELP! HURRY. ALMOST OUT OF TIME

Physics

2 answers:

IRISSAK [1]3 years ago

6 0

Answer:

balanced hurry up fam

gurry hurry hurry

Nataly [62]3 years ago

3 0

3) it’s balanced.
.......

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If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti

Ann [662]

We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km

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3 years ago

For a vehicle to negotiate a banked curve in poor weather conditions, where the force of friction f = 0, for a given velocity v

djverab [1.8K]

Answer:

$R=240m$

Explanation:

From the question we are told that:

Velocity $v=25m/s$

Force of friction f = 0

Angle $\theta=15$

Generally the equation for Radius of curvature is mathematically given by

$R=frac{v^2}{tan\theta *g}$

$R=frac{25^2}{tan 15 *9.81}$

$R=240m$

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2 years ago

Monochromatic light falls on two very narrow slits 0.048 mm apart. successive fringes on a screen 5.00 m away are 6.5 cm apart n

atroni [7]

In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
$y= \frac{m \lambda D}{d}$
where D=5.00 m is the distance of the screen from the slits, and
$d=0.048 mm=0.048 \cdot 10^{-3}m$ is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
$\lambda = \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}= 6.24 \cdot 10^{-7}m$

And from the relationship between frequency and wavelength, $c=\lambda f$ , we can find the frequency of the light:
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz$

4 0

3 years ago

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

$v=\sqrt{\frac{T}{(m/l)}}$

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

$(m/l)=\frac{10}{V^2}$

Explanation:

From the question we are told that

Speed of a transverse wave given by

$v=\sqrt{\frac{T}{(m/l)}}$

Maximum Tension is $T=10.0N$

Generally making $(m/l)$ subject from the equation mathematically we have

$v=\sqrt{\frac{T}{(m/l)}}$

$v^2=\frac{T}{(m/l)}$

$(m/l)=\frac{T}{V^2}$

$(m/l)=\frac{10}{V^2}$

Therefore the Linear mass in terms of Velocity is given by

$(m/l)=\frac{10}{V^2}$

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3 years ago

Label the producers and consumers, then label the different types of consumers.

sdas [7]

The base of the pyramid has the producers and everything else above the base falls under the consumers category i.e the locusts,frogs and the snake. The grass is the producer, the locust is a consumer, the frog is a special type of omnivore, termed the "life-history omnivore" since they eat both plants and animals but at different times in their lives. In this case they are just omnivores and lastly, the snake is a carnivore.

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3 years ago