All categories

3 years ago

a motorcycle is capable of accelerating at 5.1 m/s starting from rest how far can it travell in 1.5 sec

Physics

1 answer:

Katarina [22]3 years ago

5 0

The answer is 21m because the motion is in one dimension with constant acceleration.

The initial velocity is 0, because it started from rest, the acceleration ax is 4.7ms2, and the time t is 3.0s

Plugging in our known values, we have

Δx=(0)(3.0s)+12(4.7ms2)(3.0s)2=21m

You might be interested in

When different resistors are connected in parallel across an ideal battery, we can be certain that: a) their equivalent resistan

sergeinik [125]

Answer:

b) the potential difference across each is the same.

Explanation:

When resistors are connected in parallel with each other

then the terminals of all the resistors will connected across the terminals of the battery

So we know the potential difference of the battery is same across all the resistors

So we can say that the equivalent resistance of all the resistance is

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

so its not the mean of all resistors

also we know that the resistance are all different so power across each resistance is different given as

$P = \frac{V^2}{R}$

also current in each resistance is also different and given by

$i = \frac{V}{R}$

so correct answer will be

b) the potential difference across each is the same.

5 0

2 years ago

If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______

adell [148]

Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

$F = \dfrac{GMm}{r^2}$ ............(1)

new gravitational force

$F' = \dfrac{GMm}{r'^2}$

now from the given condition

$F' = \dfrac{GMm}{(3r)^2}$

$F' = \dfrac{GMm}{9r^2}$

$F' = \dfrac{1}{9}\dfrac{GMm}{r^2}$

now, from equation (1)

$F' = \dfrac{1}{9}F$

$\dfrac{F'}{F} = \dfrac{1}{9}$

now, the change in gravitational force factor is equal to $\dfrac{1}{9}$

Hence, the correct answer is option A

3 0

3 years ago

O((>ω< ))oo((>ω< ))oo((>ω< ))oo((>ω< ))oo((>ω< ))o

Paladinen [302]

Answer:

(•_•)

Explanation:

7 0

2 years ago

Which model shows how a comet's tail changes during its orbit?

pentagon [3]

Third model shows how a comet's tail changes during its orbit...

mark brainliest

5 0

2 years ago

Read 2 more answers

Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe

LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame $v_{m}$ 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = $\frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}$

X = $\frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}$

X = $8.07 c$

Now,

Y = y = - 2

Z = z = 3

Now,

$t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}$

$t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s$

4 0

3 years ago