Ask question

Ask question

All categories

3 years ago

14

A polar bear runs at a speed of 8.3 m/s, how far will it travel in 60 seconds?

2 answers:

Kay [80]3 years ago

6 0

498 mps since you just times 8.3 by 60

Elanso [62]3 years ago

3 0

Answer:

it would have run 498 miles

Explanation:

if the polar bear runs 8.3 every second then multiply 8.3 times sixty to get 498

You might be interested in

The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, w

jek_recluse [69]

Answer:

$XY=636N$

Explanation:

From the question we are told that:

Distance $d=0.35m$

Angle $\theta=32\textdegree$

Force $F=750N$

Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by

$XY=Fcos\theta$

$XY=750cos 32\textdegree$

$XY=636N$

4 0

3 years ago

On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

$\theta=53^{\circ}$

$\mu_k=0.16$

Speed,v=1.5m/s

Net force in x direction must be zero

$F_{net}=0$

$Fsin\theta-f=0$

$Fsin\theta=f$

Net force in y direction

$N-mg-Fcos\theta=0$

$N=mg+Fcos\theta$

$f=\mu_kN=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_kmg+\mu_kFcos\theta$

$Fsin\theta-\mu_kFcos\theta=\mu_kmg$

$F(sin\theta-\mu_kcos\theta)=\mu_kmg$

$F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}$

Power,P=Fv

$P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v$

Where $g=9.8m/s^2$

B.Substitute the values

$P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5$

$P=40.17W$

6 0

3 years ago

a mass of 10kg is placed on a horizontal table with coefficient of friction is 0.5. if the mass is static, determine weight of t

zloy xaker [14]

Answer:

10

Explanation:

5 0

3 years ago

Why might surface mining be less risky for miners than underground mining?

aev [14]

Surface miners work aboveground. (Apex) ^-^

6 0

2 years ago

Read 2 more answers

A particle R moves with a velocity of 6 m/s due East and another particle S moves at 8 m/s due South. What is the magnitude of t

cluponka [151]

Answer:

10

Explanation:

We can look at this problem as a triangle, r is the hypotenuse so if we take the square root of 6^2+8^2 we get 10

6 0

2 years ago