All categories

2 years ago

If the force on an object is in the negative direction, the work it does on the object must be:.

Physics

2 answers:

Inga [223]2 years ago

6 0

Answer:

The work could be either positive or negative, depending on the direction the object moves

Explanation:

topjm [15]2 years ago

5 0

The work done is also in the negative direction

You might be interested in

What would be new resistance if length of conductor is doubled and thickness is halved <br>

nignag [31]

Answer:

The new resistance comes out to be = 4 times of original resistance .

6 0

3 years ago

A vector R is resolved into its components, Rx and Ry. If the ratio of is 2, what is the angle that the resultant makes with the

Artemon [7]

The angle is 26.56 degrees

6 0

3 years ago

Read 2 more answers

A vehicle moving along at 5m/s. What should be the constant deceleration in order to stop it within 15m?

Oduvanchick [21]

Answer:

a = -5/6 m/s²

Explanation:

v² = u² + 2as

a = (v² - u²) / 2s

a = (0² - 5²) / 2(15)

a = -5/6 m/s²

This assumes the initial velocity was in the positive direction.

5 0

3 years ago

A conducting coil of 1850 turns is connected to agalvanometer, and the total resistance of the circuit is 45.0 ohm.The area of e

Xelga [282]

Answer:

0.459 Tesla

Explanation:

Faraday's law:

$Emf = -N\frac{\delta\phi}{\delta t}$

Φ= NAB

V = N Δ (BA) /Δt

the change in BA

was: BA = 0 because initially B was zero.

V=IR

IR = N B A /Δt

q / Δt ×R = N B A / Δt

Or: B = q R / NA = 8.87 x 10^-3 × 45.0 / 1850×4.7 x 10^-4 =

= 0.459 Tesla

3 0

4 years ago

The position of a particle moving along an x axis is given by x = 14.0t2 - 2.00t3, where x is in meters and t is in seconds. det

Mumz [18]

a) x = $14t^{2} -2t^{3}$

at t = 5s

$x = 14*5^{2} -2*5^{3} = 14*25 - 2*125 = 350 - 250 = 100 m$

b) v = $\frac{dx}{dt}$

= $28t - 6t^2$

at t = 5 s

v = $28*5-6*25 = 140 - 150 = -10 m/s$

c) a = $\frac{dv}{dt}$

= 28 - 12t

at t = 5 s

a = 28 -12*5= 28-60= -32 m/ $s^2$

d) At maximum positive coordinate velocity = 0

So, $0 = 28t - 6t^2$

$t = \frac{28}{6} = \frac{14}{3} = 4.66 s$

At t = 4.66 s

$X = 14 * 4.66^2 - 2* 4.66^3 = 202.39 m$

e) At t = 4.66 s

f) At maximum positive velocity a = 0

$0=28-12t$

$t = \frac{28}{12} = \frac{7}{3} = 2.33 s$

At t = 2.33 s

V = $28*2.33- 6*2.33^2= 32.67 m/s$

g) t = 2.33 s

h) When particle is not moving v = 0

So $0= 28t - 6 t^2$

$t = \frac{28}{6} = 4.66 seconds$

At t = 4.66 s

a = $28 - 12 * 4. 66 = -27.93m/s^2$

i) At t = 0s, X =0m

t = 5s, X = 100m

So, Displacement = 100m

Velocity = $\frac{Displacement}{Time} = \frac{100}{5} = 20m/s$

8 0

3 years ago