If two cars have the same velocity do they have the same acceleration

A student pushes a 40-N block across the floor for a distance of 10 m. How much work was applied to move the block? A. 4 J B. 40

damaskus [11]

Now for this problem, what is given is a 40 Newtons which would represent the force to be applied to the object, and a distance of 10 meters after the application of the said force. When these two combine, work is done. The unit for work is Joules and this is what we are looking for. The formula to get Joules or for work would be the force applied to the object multiplied by the distance that it travelled after the application of the force. It looks like this

work = force x distance
Joules = Newtons x meter

so let us substitute the variables to their corresponding places

Joules = 40 N x 10 m
Joules = 400 J

So the answer to this question would be C. 400 J

7 0

3 years ago

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A motor must lift a 1500-kg elevator cab. The cab's maximum occupant capacity is 400 kg, and its constant "cruising" speed is 1.

natka813 [3]

Answer:

12900 W

24200 W

Explanation:

Given:

v₀ = 0 m/s

v = 1.3 m/s

t = 2.0 s

Find: a and Δx

v = at + v₀

(1.3 m/s) = a (2.0 s) + (0 m/s)

a = 0.65 m/s²

Δx = ½ (v + v₀) t

Δx = ½ (1.3 m/s + 0 m/s) (2.0 s)

Δx = 1.3 m

While accelerating:

Newton's second law:

∑F = ma

F − mg = ma

F = m (g + a)

F = (1500 kg + 400 kg) (9.8 m/s² + 0.65 m/s²)

F = 19855 N

Power = work / time

P = W / t

P = Fd / t

P = (19855 N) (1.3 m) / (2.0 s)

P ≈ 12900 W

At constant speed:

Newton's second law:

∑F = ma

F − mg = 0

F = mg

F = (1500 kg + 400 kg) (9.8 m/s²)

F = 18620 N

Power = work / time

P = W / t

P = Fd / t

P = Fv

P = (18620 N) (1.3 m/s)

P ≈ 24200 W

5 0

4 years ago

A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh

lianna [129]

Answer: 3.21 N

$Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}$

$\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg$

For weight, we will multiply by $g=9.8 m/s^{-2}$

$weight= 0.328\times9.8=3.21\hspace{1mm}N$

Hence, the rock would weigh 3.21 N.

3 0

3 years ago

What is the velocity in meters per second of a runner who runs exactly 110 m toward the beah in 72 seconds?

rjkz [21]

B) 1.53 m/s towards the beach

8 0

4 years ago

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Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric e

kvv77 [185]

This question is incomplete, the complete question is;

A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.

Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.

Answer:

the required potential difference, between the capacitor plates is 600 V

Explanation:

Given the data in the question;

B = 0.60 T

d = 2.0 mm = 0.002 m

v = 5.0 × 10⁵ m/s.

since particle pass straight through without deflection.

F $_{net$ = 0

so, F $_E$ = F $_B$

qE = qvB

divide both sides by q

E = vB

we substitute

E = (5.0 × 10⁵) × 0.6

E = 300000 N/C

given that; potential difference ΔV = Ed

we substitute

ΔV = 300000 × 0.002

ΔV = 600 V

Therefore, the required potential difference, between the capacitor plates is 600 V

5 0

3 years ago