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Paha777 [63]
2 years ago
13

If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?

Physics
1 answer:
HACTEHA [7]2 years ago
5 0

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

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6 0
3 years ago
Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N
Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0
1 year ago
The y component of the electric field of an electromagnetic wave traveling in the +x direction through vacuum obeys the equation
Alexxx [7]

Answer: 8.6 µm

Explanation:

At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:

Ey =Emax cos (kx-ωt)

So, we can write the following equality:

ω= 2.2 1014 rad/sec

The angular frequency and the linear frequency are related as follows:

f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec

In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.

The wavelength, speed and frequency, are related by this equation:

λ = c/f

λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.

7 0
3 years ago
12. An electric motor is plugged into a standard wall socket and is run- ning at normal speed. Suddenly, some dirt prevents the
rusak2 [61]

the answer would be B. the back emf increases, and the current drawn from the socket increases

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8 0
3 years ago
A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi
malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude mg, where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

F_{net}=ma

where

F_{net} is the net force

m = 75.0 kg is the mass of the bicyclist

a=2.20 m/s^2 is its acceleration

Solving, we find the net force:

F_{net}=(75.0)(2.20)=165 N

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

F_{net}=F-R

where:

F is the forward force of push

R is the air drag

We know that:

F_{net}=165 N is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

F=F_{net}+R=165+60=225 N

6 0
3 years ago
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