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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
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A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

The mass of the meter stick is 126.99115 g

6 0
3 years ago
Read 2 more answers
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
You purchase a ten-year 1,000 bond with semiannual coupons for 982. The bond has a 1,100 redemption payment at maturity, a nomin
vladimir2022 [97]

Answer:

ummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

6 0
3 years ago
The nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports
Katarina [22]
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.

c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02 
 
Alternative solution:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Therefore, they have to fly 82.02 mph
5 0
3 years ago
Read 2 more answers
A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
3 years ago
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