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nasty-shy [4]
4 years ago
9

Will an object with more mass roll faster down a hill?

Physics
1 answer:
inna [77]4 years ago
5 0
Over time, yes. It will over time gain more momentum
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1. A limiting factor for using nuclear energy is the
Goryan [66]

Answer:

waste

Explanation:

took da test. nhgvgg

8 0
3 years ago
Read 2 more answers
A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete
ale4655 [162]

Answer:

The  last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

cos X=\frac{x^2-y^2-z^2}{-2yz}

or

cos X=\frac{y^2 + z^2-x^2}{2yz}

substituting the values in the equation we get,

cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}

or

cos X=0.8951

or

X = 26.47°

similarly,

cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}

or

cos Y=0.649

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0
3 years ago
A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the gen
Zinaida [17]

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s

N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns

Therefore, the number of turns of wire needed is 573.8 turns

4 0
3 years ago
Which electromagnetic wave enables us to see objects?
Valentin [98]
Choose visible light
3 0
3 years ago
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¿Qué distancia recorre un móvil que lleva una aceleración de 5m/s durante 10seg?
coldgirl [10]

Answer:

250 m

Explanation:

The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

Assuming the car starts from rest,

u = 0

Also we know that

a = 5 m/s^2 (acceleration of the car)

t = 10 s

Substituting, we find the distance covered:

s=0+\frac{1}{2}(5)(10)^2=250 m

3 0
3 years ago
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