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Flura [38]
3 years ago
6

If the velocity of an object is zero at one instant, what is true about the acceleration of that object

Physics
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

I. The acceleration could be positive.

II. The acceleration could be zero.

III. The acceleration could be negative.

Explanation:

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

Velocity = \frac{distance}{time}

If the velocity of an object is zero at one instant, then the acceleration of that object may have the following characteristics;

I. The acceleration could be positive.

II. The acceleration could be zero.

III. The acceleration could be negative.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is final and initial velocity respectively, measured in ms^{-1}

t is time measured in seconds.

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Doss [256]

Answer: He died later in 78 BC and was accorded a state funeral. Hearing of Sulla's death, Caesar felt safe enough to return to Rome. Lacking means since his inheritance was confiscated, he acquired a modest house in the Subura, a lower-class neighbourhood of Rome. hope this helps. Can u give me brainliest

Explanation:

3 0
3 years ago
The image produced by an object is –10.0 cm from a concave mirror that has a focal length of 5.0 cm. The distance of the object
Gennadij [26K]
Given:
Focal length: 5.0 cm
image produced: -10.0 cm

mirror equation:   1/di + 1/do = 1/f

Look for the actual distance of the object from the mirror. Solve for do.

1/do + 1/(-10) = 1/5
1/do = 1/5 + 1/10
1/do = (1/5 * 2/2) + (1/10 * 1/1)
1/do = 2/10 + 1/10
1/do = 3/10
10 = 3do
10/3 = do
3.33 = do

The distance of the object from the mirror, rounded to the nearest centimeter is 3 cm.
3 0
4 years ago
Read 2 more answers
A small metal sphere has a mass of 0.11 g and a charge of -21.0 nC . It is 10 cm directly above an identical sphere with the sam
GrogVix [38]

Answer:

A. the magnitude of the force between the spheres is 3.97 x 10⁻⁴ N

B. the magnitude of its initial acceleration is 5.83 m/s²

Explanation:

given information:

metal sphere's mass, m = 0.1 g = 1 x 10⁻⁴ kg

charge, q = -21 nC = -2.1 x 10⁻⁸

r = 10 cm = 0.1 m

What is the magnitude of the force between the spheres?

F₁₂ = k q₁q₂/r²

     = ( 9 x 10⁹) (-2.1 x 10⁻⁸)²/(0.1)²

     = 3.97 x 10⁻⁴ N

If the upper sphere is released, it will begin to fall. What is the magnitude of its initial acceleration?

mg - F₁₂ = ma

a = g - (F₁₂/m)

   = 9.8 - (3.97 x 10⁻⁴/1 x 10⁻⁴)

   = 5.83 m/s²

5 0
3 years ago
A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.
strojnjashka [21]

Answer:

D=99.4665307m \approx 99.5m

Explanation:

From the question we are told that

Mass  m=6kg

Velocity of mass  V_m=16

Force of Tunnel  F_t=8N

Length of Tunnel L_t=1.6

Height of frictional incline H_i=2.9

Angle of inclination  \angle =16 \textdegree

Acceleration due to gravity  g=9.8m/s^2

First Frictional surface has a coefficient  \alpha_1 =0.21\ for\ d_c=1

Second Frictional surface has a coefficient \alpha _2=0.1

Generally the initial Kinetic energy is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}(6)(16)^2

K.E=768

Generally the work done by the Tunnel is mathematically given as

w_t=F_t*d_t

w_t=8*1.6

w_t=12.8J

Therefore

Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t

E_t=K.E+E_t\\E_t=768J+12.8J

E_t=780.8J

Generally the energy lost while climbing is mathematically given as

E_c=mgh

E_c=(6)(9.8)(2.9)

E_c=170.52J

Generally the energy lost to friction is mathematically given as

E_f=\alpha *m*g*cos\textdegree*d_c

E_f=0.21*6*9.8*cos16*1

E_f=11.86965942 \approx 12J

Generally the energy left in the form of mass Em is mathematically given as

E_m=E_t+E_c+E_f

E_m=(768J)-(170.52)-(12)

E_m=585.48J

Since

E_m=\alpha_2*g*m*d

Therefore

It slide along the second frictional region

D=\frac{585.46}{0.1*9.81*6}

D=99.4665307m \approx 99.5m

6 0
3 years ago
I need help with question two
zaharov [31]
I can't see numbers here, so here are all the answers:
1) the frequency is c/λ = 3e8/556e-9 = 5.39e14Hz
2) light travels at <span>299,792,458 m/s.  So in nanoseconds it's 0.299792458m.  This is about 1/3 of a meter which is about one foot.
3) length is L = ct = (</span>299,792,458 m/s)(6e-15) = 1.799e-6m or 1.799μm

6 0
3 years ago
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