Ask question

Ask question

All categories

3 years ago

6

A boy pushes forward a cart of groceries with a total mass of 40.0 kg. What is the acceleration of the cart if the net force on

the cart is 60.0 N?

1 answer:

melisa1 [442]3 years ago

6 0

1.5m/s2

F=ma

a=F/m

a=60/40

a=1.5m/s2

You might be interested in

Given the picture below, what is the volume of the object?

Valentin [98]

The answer is D, 10 ml

4 0

3 years ago

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0

3 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Brrunno [24]

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector $X = 7.8*Cos(50) = 5.01 Km$ y $y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km)$ so that $Tan (\alpha ) = \frac{0.375}{5.01} = 4.28$ ; $\alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree)$ to get how far we use Pythagorean theorem so $R^{2} = x^{2}+y^{2}$ so that $R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)$

6 0

3 years ago

Read 2 more answers

An airplane with an airspeed of 120 km/h has a heading of 30 degree east of North in a wind that is blowing toward the east at 6

lesya692 [45]

Answer:

Explanation:

Velocity of plane relative to ground V_pg = ?

Given the velocity in vector form ,

velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j

V_wg = 60 i

V_pg = V_pw +V_wg

= 120 cos30 i + 120sin30j + 60i

= 164 i + 60 j

magnitude

=251 km / h

=

8 0

3 years ago

Which statement describes something that should always be done at the end of a calculation?

Reika [66]

Answer:

C) Check that the numerical answer is reasonable and the units are what you expected.

Explanation:

5 0

3 years ago

Read 2 more answers