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3 years ago

A racecar begins at rest and accelerates to 25 m/s at a rate of 6.25 m/s2. What distance does the racecar cover?

Physics

1 answer:

Vitek1552 [10]3 years ago

7 0

Answer:

50 m

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s =(25² - 0²) / (2(6.25))

s = 50 m

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A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 mm behind the nose of the airplane, the front w

padilas [110]

Answer:

The percentage of the weight supported by the front wheel is A= 19.82 %

Explanation:

B] Let the mass of plane be m, force on backwheels be Nb and nosewheel be Nn

Torque equation about nose wheel,

mg*(2.58-0.8) - Nb *(3.02-0.8) = 0

Nb = mg*(2.58-0.8)/(3.02-0.8) = 0.8018 mg

Nn = mg - Nb = (1-0.8018) mg = 0.1982mg

Weight percentage supported by front wheel = 19.82% answer

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3 years ago

Which of the following are heterogeneous mixtures (not homogeneous mixtures)?

BARSIC [14]

The following are heterogeneous mixture: MILK THAT IS SOLD AT THE STORE, A CAKE MIX AND MIXED NUT.An heterogeneous mixture is one which contained many parts which can be separated out.A smoothie is an homogeneous mixture because the juice of different fruits that are use can not be separated out.
Read more on Brainly.com - brainly.com/question/4731209#readmore

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3 years ago

As an astronaunt travels from the surface of the earth to a postion that is four times

allochka39001 [22]

Answer:

Explanation:

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3 years ago

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A block is on a frictionless table, on earth. The block accelerates at 7.5 m/s when a 70 N horizontal force is applied to it. Th

Liono4ka [1.6K]

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, $m=\dfrac{F}{a}$

$m=\dfrac{70\ N}{7.5\ m/s^2}$

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

$W=m\times g$

$W=9.34\ kg\times 1.62\ m/s^2$

W = 15.13 N

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

3 0

3 years ago

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

$\boxed{B = 0 T}$

3 0

2 years ago