All categories

3 years ago

A commuter backs her car out of her garage with an acceleration of . (a) How long does it take her to reach a speed of 2.00 m/s

1 answer:

3 0

Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s

Answer:

1.43 s

Explanation:

Applying,

a = (v-u)/t........... Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t = time

make t the subject of the equation

t = (v-u)/a........... Equation 2

From the question,

Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²

Substitute into equation 2

t = (2-0)/1.4

t = 1.43 s

You might be interested in

A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon

jek_recluse [69]

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

Στ = 0

W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

x₂ = $\frac{M_1g }{m_2 g} \ x_1$

let's calculate

x₂ = $\frac{100}{75} \ 1$

x₂ = 1.33 m

7 0

3 years ago

Can you help and explain these for me?

ollegr [7]

Yes I can do you want me to

8 0

3 years ago

A ferry approaches shore, moving north with a speed of 6.2 m/s relative to the dock. A person on the ferry walks from one side o

jasenka [17]

Speed of Ferry is towards North with magnitude 6.2 m/s

Here if we assume that North direction is along Y axis and East is along X axis then we can say

$\vec v_f = 6.2 \hat j$

Now a person walk on ferry with speed 1.5 m/s towards east with respect to Ferry

so it is given as

$\vec v_{pf} = 1.5 \hat i$

also by the concept of relative motion we know that

$\vec v_{pf} = \vec v_p - \vec v_f$

now plug in all values in it

$1.5 \hat i = \vec v_p - 6.2 \hat j$

$\vec v_p = 1.5 \hat i + 6.2 \hat j$

now if we need to find the speed of the person then we need to find its magnitude

so it is given as

$v = \sqrt{1.5^2 + 6.2^2}$

$v = 6.37 m/s$

7 0

3 years ago

How does the transfer of thermal energy occur?

natali 33 [55]

Heat<span> may be </span>transferred<span> by means of conduction, convection, or radiation. </span>

8 0

3 years ago

A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it

mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0

3 years ago