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9 months ago

Which of acceleration, age, speed, temperature, and velocity are vector quantities?.

Physics

1 answer:

ollegr [7]9 months ago

4 0

Velocity and acceleration are vector quantities whereas speed, temperature and age are not.

<h3>What is a vector quantity?</h3>

Vector is a quantity that has both magnitude and direction and is represented by an arrow whose direction is same as that of the quantity and length is proportional to the quantity's magnitude.

Vector has magnitude and direction but it does not have position. Velocity and acceleration both are vector quantities as they have magnitude and direction.

If the speed of an object remains same but direction changes then the object is accelerating. It is important to remember that acceleration and velocity aren't always in the same direction.

To know more about vector quantity, refer

brainly.com/question/626479

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1. Your new designer chair has an S-shaped tubular metal frame that behaves just like a spring. When your friend, who weighs 600

Anvisha [2.4K]

Explanation:

Given that,

Weight of the friend, W = 600 N

When the friend sits on the metal frame it bends downward 4 cm, we can say that the compression in the it is 4 cm or 0.04 m

To find,

Spring constant for this chair or k

Solve :

The weight of an object is equal to the force exerted by the gravitational force, F = 600 N

According to Hooke's law, the force exerted by the spring is given by :

F = kx

k is the spring constant

$k=\dfrac{F}{x}$

$k=\dfrac{600\ N}{0.04\ m}$

k = 15000 N/m

Therefore, the spring constant of the spring is 15000 N/m.

5 0

3 years ago

The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance

belka [17]

The gravitational force between the spheres is

$F_{\rm g}=\dfrac{G(188\,\mathrm{kg})(0.93\,\mathrm{kg})}{(0.27\,\mathrm m)^2}\approx1.6\times10^{-7}\,\mathrm N$

where <em>G</em> = 6.674 x 10⁻¹¹ N m²/kg².

The weight of the lighter sphere is

$F_{\rm w}=(0.93\,\mathrm{kg})g\approx9.1\,\mathrm N$

where <em>g</em> = 9.80 m/s².

The ratio between the two forces is then

$\dfrac{F_{\rm g}}{F_{\rm w}}\approx1.8\times10^{-8}$

7 0

3 years ago

At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?

Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$$m / s^2$

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$$m/s^2$ is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}$

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

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4 0

1 year ago

A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

Kisachek [45]

Answer:

$B_2 = 1.71 mT$

Explanation:

As we know that the magnetic field near the center of solenoid is given as

$B = \frac{\mu_0 N i}{L}$

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

$\frac{B_1}{B_2} = \frac{L_2}{L_1}$

now plug in all values in it

$\frac{2.0 mT}{B_2} = \frac{21}{18}$

$B_2 = 1.71 mT$

3 0

2 years ago

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

8 0

3 years ago