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Zielflug [23.3K]
4 years ago
10

If a rock is thrown upward on the planet Mars with a velocity of 10 ms, its height in meters seconds later is given by (a) Find

the average velocity over the given time intervals: (i) [1, 2] (ii) [1, 1.5]
Physics
1 answer:
Delicious77 [7]4 years ago
6 0

Answer:

1) X(t)=Xo+So(t)+\frac{1}{2} g(t)^{2}

2) S(Δt)=So+g(Δt)

Explanation:

I think the equation they gave you for the Height upon time in seconds it's (1), where if you see, you will find the gravity, that you should multiply by 2 because, its divided by two in (1) (that should be your (a)), then, once you find your gravity, you can use the equation (2) to know the Final Speed replacing g , at the time asked, remember that g is gravity, and Δt is the: final time- initial time.

so in the [1,2] interval of time, your Δt=1s, and in [1,1.5] is Δt=0,5s.

i hoped it helped you even though i cant give you the exact answer right now.

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A 0.250 kg fan cart accelerates at 24 cm/ s^2 for 4.5 seconds. what is the net force (fan thrust minus drag and friction) on the
Norma-Jean [14]

Answer:

0.06 N

1.08 m/s

Explanation:

m = mass of the fan cart = 0.250 kg

a = acceleration of the fan cart = 24 cm/s² = 0.24 m/s²

F = Net force on the cart

Net force on the cart is given as

F = ma

F = (0.250) (0.24)

F = 0.06 N

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart

t = time interval = 4.5 s

Using the equation

v = v₀ + a t

v = 0 + (0.24) (4.5)

v = 1.08 m/s

6 0
3 years ago
What is the resistance of : A) A 1.70 m long copper wire that is 0.700 mm in diameter? B) A 20.0 cm long piece of carbon with a
astra-53 [7]

Answer:

(I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

Explanation:

Given that,

Length of copper wire = 1.70 m

Diameter = 0.700 mm

Length of carbon piece = 20.0 cm

Cross section areaA = (2.00\times10^{-3})^2\ m

(I). We need to calculate the area of copper wire

Using formula of area

A=\pi r^2

A=3.14\times(\dfrac{0.700\times10^{-3}}{2})^2

We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{1.68\times10^{-8}\times1.70}{3.14\times(\dfrac{0.700\times10^{-3}}{2})^2}

R=0.0742\ \Omega

(II). We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{3.5\times10^{-5}\times20\times10^{-2}}{(2.00\times10^{-3})^2}

R=1.75\ \Omega

Hence, (I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

8 0
3 years ago
A block of mass 10.0 kg is pulled to the right along a rough horizontal surface with a constant horizontal force of 20.0 N. The
Yakvenalex [24]

Answer:

the magnitude of acceleration will be 1.50m/s^2

Explanation:

To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma

if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.

The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).

you will find that Fnet=15.0N

With that, plug in the values you know to calculate the acceleration of the block:

Fnet=ma

(15.0N)=(10.0kg)a    from her you can divide both sides by 10 to isolate a:

1.50=a  (and now make sure to label the units of your answer)

a=1.50m/s^2 (which is the typical unit for acceleration)

7 0
3 years ago
An electron moves with velocity v⃗ =(5.9i−6.4j)×104m/s in a magnetic field B⃗ =(−0.63i+0.65j)T. Determine the z-component of the
HACTEHA [7]

Answer:

Explanation:

Force on the electron = q ( v x B )

q = - 1.6 x 10⁻¹⁹

v = (5.9i−6.4j)×10⁴

B = (−0.63i+0.65j)

v x B = (5.9i−6.4j)×10⁴  x (−0.63i+0.65j)

= (3.835  - 4.032 ) x 10⁴ k

= - 1970 k

Force on the electron = q ( v x B )

= - 1.6 x 10⁻¹⁹ x -1970 k

= 3.152 x 10⁻¹⁶ k

z-component of the force on the electron

Fz = 3.152 x 10⁻¹⁶ N  

7 0
3 years ago
Imagine a box sitting on a shelf. What forces are acting on the box?
mel-nik [20]

In case of an object sitting at rest on another base, there are two equal and opposite forces – Normal force and the gravity.

Answer: Option A

<u>Explanation: </u>

When an object is placed at rest position on another object, there is a force exerted by the surfaces of the two contact objects. This force is denoted as Normal Force.

When an object such as a box is placed on a shelf, its surface exerts a contact force on the base of the shelf- The Normal force directed upward. Meanwhile, the gravity stays at its action and tries to pull the box towards itself.  

Both of these forces however are equal and opposite and therefore, there is zero net force on the box. That's why it remains at rest, holding on Newton's third law.

8 0
3 years ago
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