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sp2606 [1]
3 years ago
9

Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effecti

ve in pulling the sled horizontally along the ground?
Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0
If we consider that the surface wherein this box is located is frictionless in such a way that the friction does not affect its motion, we can calculate for the horizontal force by multiplying the force with the sine of the angle.
                     x-component of force = (4 N) x (sin 62°)
                                         = 3.53 N
You might be interested in
A rocket moves upward, starting from rest with an acceleration of +30.0 m/s2 for 5.00 s. It runs out of fuel at the end of this
olchik [2.2K]

Answer:

The distance covered by the rocket after fuel ran out is 3442.04 m

Explanation:

Given that the rocket moves with an acceleration a=30m/s^2

time t=5 s

Since the rocket starts from rest initial velocity  u=0 s

The distance it travelled within this time is given by  s=ut+ \frac{1}{2} at^2                                                                                                  =0 \times 5+ \frac{1}{2} (30\times25)=375 m

Velocity at this point is given by v=u+at

v=0+30\times5=150m/s

Given that at this height it runs out of fuel but travels further. Here final velocity v=0(maximum height), initial velocityu=150 m/s  and time to zero velocity t=\frac{v}{g} = \frac{150}{9.8} =15.3 s.

Thus it travels 15.3 seconds more after fuel running out. The distance covered during this period is given

s= ut+\frac{1}{2} gt^2=150 \times 15.3+1/2 \times9.8 \times 15.3^2=3442.04 m

7 0
3 years ago
Gold has a density of 19.3 g/cm3. What is the mass of a 5 cm3 block of gold?
Tems11 [23]
Do not forget that mass = <span>volume x density
</span>Mass of 1 cm^3 = Density[/tex]
mass of 2 cm^3 = 19.3 g + 19.3 g = 2*19.3 g
Then eventually we can find <span>mass of 5 cm^3 : = 
</span>19.3 g + 19.3 g+19.3g+19.3g+19.3g= 5*19.3 g
So the answer is D
<span>And that's it. I'm sure it will help.</span>
5 0
3 years ago
The brakes application to a car produce an acceleration of 6ms2 in the opposite direction to the motion .If the car takes 2 seco
Anna [14]

Answer:12 meter

Explanation:

acceleration(a)=6m/s^2

Time(t)=2 seconds

Distance =(a x t^2)/2

Distance =(6 x 2^2)/2

Distance=(6 x 2 x 2)/2

Distance=24/2

Distance =12

Distance is 12 meters

7 0
2 years ago
A bus hits a bug and the bug splatters on the windshield, which force is greater?
grandymaker [24]

The forces acting on a body and the type of motion that results are given by Newton's three Laws of motion

  • The <u>force </u>of the bus is <u>the same</u> as reactive force of the bug

Reason:

According to Newton's third Law of motion, given that the bug collides

with the bus, the force with which the bus hits the bug is equal to the

reactive force of the bug on the on the bus

According to Newton's second Law of motion, force is equal to the rate of

change of the momentum produced

The impulse of the force of the bus on the bug is given as follows;

F·Δt = m·(v₂ - v₁)

Given that the force of the bus is large, the change in momentum, m·(v₂ - v₁),

is also large such that the parts of the bug are split by the rapid change in

velocity, and the bug splatters on the windshield, and is then carried along

on the trip,

The equally large reactive force of the bug, is such that the bug splatters

due its magnitude

Therefore, the correct response is that <u>the forces are the same</u>

Lean more here:

brainly.com/question/21279060

8 0
2 years ago
Question 20
oksian1 [2.3K]

Answer:

The image distance is 17.56 cm

Explanation:

We have,

Height of light bulb is 3 cm.

The light bulb is placed at a distance of 50 cm. It means object distance is, u =-50 cm

Focal length of the lens, f = +13 cm

Let v is distance between image and the lens. Using lens formula :

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{13}+\dfrac{1}{(-50)}\\\\v=17.56\ cm

So, the image distance is 17.56 cm.

5 0
3 years ago
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